 Hi guys, how are you there? Hello? Are you able to hear me? Yes Yeah, who are there? Gurman. I can see Chitij. Chitij, you're there? Yes And who is Al-Sarang and Swamiya? Hmm. So we were doing inequalities Isn't it? So could you get a chance to solve those questions guys? You could you could solve any of them? Hello, tell me. Yes. No, don't know How many could you solve Gurman? I didn't get four of them. Four of out of ten. How many did I post? Sir six. No More than that. Eleven. Eleven Eleven. So you could solve Yeah, Swamiya, how many could you solve? Are you able to see the screen now guys? You solved four. Okay, and what about others? Chitij, how many? Sarang? Sarang? Sir four. Four you could solve? Yeah. You're tempted. There's a difference between solving and attempting. No, I solved. You solved. And Sarang? And not through the cheat methods. Not through cheat methods? Yeah. What about others? Who else is there? I can see Gurman only, Chitij and Sarang. Sarang is there. Sarang, you're there? Can you hear me? Sir, by mistake he left the call. Oh, okay. Never mind. So he'll join. So let's carry on guys and Those who have done quadratic polynomials, they would be aware of something called discriminant, isn't it? So today we will have this another session on inequalities where we will learn some more Inequalities and the theorems, right? So first is today's this thing is quadratic inequality. Okay, so Quadratic inequality. So for those questions you could not solve, you can tell me the numbers later on. I'll send you the solutions. Okay, so quadratic inequality. What do we know? We know that quadratic equation is of the form of a x square plus b x plus c equals to 0, isn't it? Where a, b, and c all are real numbers. All are real numbers and a is not equal to 0. And for this quadratic equation to have real roots, what do we know? We know that discriminant d is equal to b square minus 4ac. Right? This discriminant must be greater than equal to 0 for real and unequal roots. Okay. And what else? Let us say if I have a polynomial like y is equal to a x square plus b x plus c. Okay, so hence y is always greater than equal to 0 if discriminant is less than equal to 0. Okay, so this is property of let us say a quadratic polynomial. Let us say this is my x and I'm drawing the x and y axis. This is x. This is y. Okay, and we had let us say this equation. This polynomial has this kind of a this kind of an let's say graphical representation. So if you see what is this? This particular curve is y is equal to y. Yeah, y is equal to a x square plus b x plus c. So if this polynomial is greater than equal to 0, the condition would be that discriminant must be less than equal to 0. Okay, so basis this, this is a you know, and let us say if the curve is something like that. So most of you I have taught already in the class that for this case a is greater than 0 and for this case a must be less than 0. Okay, so now you can see in this case also if there is a if there is a quadratic polynomial whose a is less than 0 and discriminant is also less than 0, then this particular y will always be less than 0. Can you see this y is always less than 0? In this case y is greater than 0. Right, so these are all inequalities, right? So hence let us solve one problem. The problem is given that question. Try to solve this that a is greater than equal to b is greater than equal to c. Let us say this is given, right? Where a, b and c all are real numbers. Now prove that, prove that a square plus ac plus c square is greater than equal to price of b and then a minus b plus c. Can you try to prove this? Try for two minutes and let me see if you can prove it. Did you get the question? So a, b and c are real numbers such that a is greater than equal to b is greater than equal to c. Prove that a square plus ac plus c square is greater than equal to 3b times a minus b plus c. Hint is you have to use quadratic inequality. Any breakthrough? Yes guys, any. So another hint is try to write this as a quadratic in one of the variables and then try to find out the b and see what happens. Another hint. Try to, as I told you, express in terms of one variable and form a quadratic equation and then. Gurman, why? Done? No, sir. Why are you clicking so much so many times? Sir, there was a notification, so I was just switching it off. So now, okay, anyone, you could do it. Try to find out, try to frame a quadratic equation in one of the variables. Do you see a quadratic equation here? Yeah. Yeah. Who else is there? I can see Sarang has joined. Sarang is there? Yes, sir. Hello. Are you there? Could you try this one? Yeah. Any breakthrough in this question? Sir, I'm trying it, sir. See, in such cases guys, it was clearly evident that you had to use something called quadratic inequality. What is that? So let me just rewrite this inequality. It is a square plus ac plus c square, okay? And this is greater than equal to what? It is 3ab minus 3b square plus 3bc. Is it it? Now, let us write this such a way that all the terms are in one, on one side. So hence, I can write this as a square plus ac plus c square plus 3b square minus 3ab and minus 3bc. And this must be greater than equal to 0. So if you see, there are, which term should I go for? And you have square term in a as well, square term in b as well, and square term in c as well. So but if you see, b has more, for that matter, you could have taken any of the three as, but a square b square c square all are there. So what I can do is, let me rewrite in terms of b square. So 3b square, then what? I can take minus 3b as common and this is a plus c. And what is left? Left is plus a square plus ac plus c square and this must be equal to 0. This is what it is left with, isn't it? So if you treat it as a quadratic in b, so discriminant d will be b square. What is b square here guys? Minus 3a plus c. This is b here in this case, b square minus 4 into 3b square. Okay. Into, what's next? 3b square into a square plus ac plus c square. You get this? This is the discriminant. Now, let us try and calculate this. So this is what they got. I'm sorry, come again. Sorry, I couldn't get you Sarang. What did you say? So this is this minus 12b square times a square plus ac plus c square. Okay. So hence, this is nothing but 9 times a square plus c square, then what? Plus 2ac minus 12a square b square minus 12 ac b square and minus 12b square c square. Okay. So opening this up, we'll get 9a square plus 9c square plus 18ac. Am I right? Looks like. Sorry, b square will not be there. I'm sorry. b square is, yeah, this should not be there. Yes. So hence, b square is gone. b square is gone. Yeah, correct. So this is the final one. So hence, this is minus 12a square minus 12 ac minus 12c square. So this means what is left? Minus 3a square is left. Then what minus 3c square would be left? Then what? Plus 6ac is left. If I take 3 common, I'll get a minus 3 common, you'll get a square plus c square minus 2ac, right? Which is nothing but minus 3a minus c whole square, right? And which is clearly less than 0. Less than equal to 0, less than equal to 0. Why equal to? Because a equal to b equal to c, right? So, right, this is less than 0. So discriminant is less than equal to 0. That means what? The expression must be greater than equal to 0. And hence, this expression is true. Okay. Did you understand? So if there is a quadratic polynomial, if the d is less than equal to 0, then the polynomial itself will be greater than equal to 0. This is what was the learning from this inequality, right? So moving ahead. Next type of inequality is very famous one. And you know, most of you would be knowing, and this is AM, GM, HM. AM is always greater than equal to GM is always greater than equal to HM, right? So hence, we call it arithmetic, arithmetic mean, arithmetic mean, mean of anything. We'll define all of them is greater than geometric mean, geometric mean, mean is greater than equal to harmonic mean. And equality holds when numbers are equal to harmonic mean. Okay. So what is AM of two numbers? AM of A1 and A2 is nothing but, what is AM? Nothing but A1 plus A2 by 2. Okay. This is for two numbers. Okay. In GM, geometric mean of A1, A2 will be, A1, A2 will be nothing but A1, A2 multiplied whole to the power half, which is nothing but root of A1, A2. This is called geometric mean. And harmonic mean is, harmonic mean of A1, A2 is given by, harmonic mean is given by 1 upon 2, 1 upon A1 plus 1 upon A2. And this can be generalized for, generalized for many numbers as well, more than two numbers as well. So hence, what is it? Please remember, this is a very famous and very effective tool. AM is greater than equal to GM is greater than. HM. Okay. And let us say A1, A2, now generalizing for more numbers, A3, so on and so forth, AN all belongs to the set of real numbers. Then my friend, then A1 plus A2 plus A3 plus plus AN whole divided by N. This is the average or the arithmetic mean is greater than equal to A1, A2, A3 dot, dot, AN whole to the power 1 upon N and its root of this is greater than equal to 1 upon N, 1 upon 1 by A1, I am sorry, not 1 upon A1. This is 1 upon N, 1 by A1 plus 1 by A2, 1 by A2. I am sorry, this quantity is H. So hence, hold on, what times I am trying to, HM of A2 is, this is 1 upon, this is HM of A1, A2. Okay, 1 upon, right? So HM of A, so basically 1 upon HM, harmonic mean is nothing but 1 upon N or in this case, 1 upon 2, 2, 1 upon 2 and this is 1 by A1 plus 1 by A2. This is 1 upon HM. So you can find out what is HM? HM will be nothing but 1 upon this 4, isn't it? So hence, that will be nothing but 2 by, 2 by 1 by A1 plus 1 by A2 in this case. So here in this case will be, this one is nothing but N upon 1 by A1 plus 1 by A2 plus dot dot 1 by A. So please remember these inequalities. Okay, so now let us see some, and there are some more results based on AM, GM and HM inequality. What are these? We will also learn that. So you can take down, these are very important inequalities. So A1, sorry not A1, let us say A and B. So I am saying A1 or A2 plus B2 plus AB will be greater than equal to 3 by 4 A plus B whole square. Okay, this is, this is one of the inequality. Okay, this is also called Sophie inequality. Sophie inequality. Same person made this, Sophie is your main thing. Yeah. Okay, so you know, you have to take down these inequalities and keep them, so that you can solve the questions a little quickly. So A square plus B square again, another equality is minus AB is always greater than equal to AB. This is very easy to prove, why? Because A square plus B square minus 2 AB is always greater than 0. Why? Because A minus B whole square is always greater than equal to 0. So it follows directly from here. Then A cube plus B cube is greater than equal to AB A plus B. Can you prove this? Similarly, let me just mention all of them. One, two, three, four. Fourth is AB by A plus E is less than equal to A plus B by 4. This also is easy. Why? Because A plus B whole square is greater than equal to 0. So what can I say? Minus 4 AB right now. So A square plus B square plus 2 AB plus 4 AB. Will it be greater than equal to 0? No, minus, hold on. How do you prove this? Can you prove this one? AB, so basically you have to prove that. What do you need to prove? You need to prove that A plus B whole square is greater than equal to 4 AB. That means A square plus B square plus 2 AB minus 4 AB is greater than equal to 0. Which is true. Why? Because this is nothing but A square plus B square minus 2 AB is greater than equal to 0. Because A minus B whole square also is greater than equal to 0. So this holds. If this holds, then this holds. Correct. Then one more. There are actually there are lots, lot many more these kind. Let us say this is A square plus B square by A plus B is greater than equal to A plus B by 2. And likewise many, many more. So now that you know the AMGM inequality and their varieties, let us delve into problem solving. Here is the first problem on this. Prove that or show that question. Question is if A, B, C, V are four positive real numbers A, B, C, V all belong to R and A, B, C, V all are greater than 0. Okay. Then prove that A by B plus B by C plus C by D plus D by A is greater than equal to 4. Prove. Direct application. You know you gave this in the sheet. Did I? Yeah. So did you arrive at the solution? I did not do this. Now try. The right application guys, why are you doing so much of time? Hello. Not done. Yes. So just a minute sir. Done? Yeah. All of you. How many? Okay. So you see very important that all of them are greater than 0. So hence clearly A by B is greater than 0 and B by C is also greater than 0. So C by D is also greater than 0 and D by A also greater than 0. Okay. So hence A M, A M of 4 number will be greater than equal to GM. So what is A M of all this? A M of all this will be A by B plus B by C plus C by D plus D by A sum of all of the 4 divided by 4. This is the arithmetic mean of this. This must be greater than equal to the GM. GM is nothing but what? Multiplication of all. So A by B into B by C into C by D into D by A whole the power 1 upon 4. All right. So hence what do we get? Very easily A by B plus B by C plus E by D plus D by A by 4 is greater than equal to 1. Hence A by B plus B by C plus C by D plus D by A is greater than equal to 4. Simplified it. Simplified it. What did you do after simplification? I bought the 4 down. In the end it comes to the same thing. You bought the 4 down. So you simplify it and then you get everything something value divided by A B C D. So you bring A B C D to the side of 4 and you bring 4 down. And then it just then you can see that if you multiply all the numbers in the numerator on the left hand side and then find 4th root it's the geometric mean. I couldn't get your solution So basically simplify everything. Simplify means what? Take LCM and all that? Yeah, that's what I did. Okay, okay. Fair enough. So you are saying simplify everything. So let's simplify. So you will be getting what? A square D C plus B square AD plus B square AD B A plus C square AB. C square you'll get? Plus C square AB. C square AB okay. Plus B square B square AB. Yeah, divided by AB. Okay. Is greater than equal to 4. Right. So then like you bring A, B, C, D on right hand side and 4 on left hand side. So what you do is you take this. This is what you mean? Yeah. So now if you multiply all the terms in numerator on left hand side then find the 4th root it's A, B, C, D. So that way also it works. I see. Okay, good. So what are you saying is there are 4 terms this, this, this, this. So arithmetic of arithmetic mean of that will be equal to or greater than equal to A square CD and then there are how many A square two more. So it all will be having if you multiply all you will get A4, B4, C4, D4 and this to be power one by four, which is ABCD. Right. Good. Good. Great. So let me now give you another one. I have given you lot many in the sheet. Lot of, lot of them will be used to solve. So solve this question. Question is if X and Y are positive real numbers or let us say A and B belongs to R plus. Okay, real numbers. Then prove that 4A to the power 4 plus 4B to the power 4 plus 5A square plus B plus 1 is greater than 12AB. Yeah, prove this. 4A to the power 4 plus 4B to the power 4 plus 5A square plus A plus B plus 1 has to be more than 12AB. Again, hint is use AMGM. Sir, is it 4A cube or 4A to the power 4? A to the power 4 and then I'm sorry, this is B cube. Did I give this question to you? No, I'm just asking because then it kind of makes sense. I thought it was 4A cube. No, 4A to the power 4 plus 4B cube plus 5A square plus B plus 1. There's a pattern if you notice AB, AB. I just need to need to see your thing. Think for some time and let me know. Any breakthrough? Yes, Chutish, what are you thinking? Guys, you're silent. Yes, sir. Any thought process on this? Sir, I'm trying to find the GM and the HM. GM and HM of what? Why? GM of what? And there are lots of pluses. So why are you not going for AM? Yes, other people, Sarang, are there or not? Hello? Chutish? Yes, sir. Anything? Any breakthrough? No, sir. Let me solve now guys. Now, could you solve? Anyway, see again how to apply AMGM here. So we will see this is one thing which is A, this one and this one is going to be little, we'll club them and this one and this one is going to be clubbed together. Let us see how does it work. So if you see 4A to the power 4, plus 1, 4A to the power 4 plus 1 is certainly greater than equal to, let us say, this by 2 is greater than 4A4 into 1 square root. Is it it? Which is equal, this one is nothing but 4A to the power 4 plus 1 is greater than 2 times 2A. So hence we can say 4, sorry, 2A square. So 4A to the power 4 plus 1 is greater than 4A square. Can you see something now? So this is 1 by AMGM in inequality. Similarly, if I say 4B cube plus B, 4B cube plus B by 2 is going to be greater than equal to 4B cube into B square root. So hence it is 4B cube plus B by 2 is greater than equal to 2B square. So hence 4B cube plus B is greater than equal to 4B square. Is it okay? Guys, so that means the given expression, the given expression was 4A4 plus 4B cube plus 5, what was it? 5A square plus V, 5A square plus V plus 1 can be written as greater than equal to. So now if you club these two, there is 5A square here. So I can take out 4 from there. Is it in? So hence I can write that this is equal to sorry. So if you see these two together, so this is greater than equal to what? This is greater than equal to 4A square and 4B cube and B is greater than equal to 4B square and 5A square is left alone. So hence it is nothing but this is equal to 9A square plus 4B square, isn't it? This is 9A square plus B. Now again if you see 9A square plus 4B square by 2 will be greater than equal to 9A square into 4B square root. Which is nothing but 9 for the 36 or 6AB. So hence what do I get? 9A square plus 4B square is greater than equal to 12AB. Correct? This is what? 12AB. This is what you have to prove, right? So this entity here is eventually greater than 12AB. That is what you are supposed to find prove. So hence prove. So you have to be, you know, this is where is the conventional that is what you do in school, right? It is unconventional. So you have to think, obviously it will take some time so don't lose heart but then you have to spend time and explore, explore a lot. So I'll be again, I've already given you some questions. Now I'll be moving further and now I'm going to give you another theorem which is called weighted means. So weighted means you have to keep, you know, all these theorems and concepts at one place to be able to apply all these, okay? So given any n positive real numbers, so there are n positives real numbers, okay? There are n positive real numbers then A1, A2, let us say these are all real positive real numbers, please understand, right? And let us say W1, W2, W3 or positive weights. What is positive weights? I'll tell you positive weights. So for example, in your exam, in your, you know, 10th board you have some, let us say 20% weightages of internals and 80% is of the exam. So these are called weightages, 0.2 and 0.8 are the weightages. So let us say you got 50 out of 100 in internals and let us say 80 out of 100 in the final paper. So hence your total score will be 20% of 50, that is 20% of 50, okay? And 80% of 80 and this is out of 200, this is out of 100. So hence how much is it? So 20% of 50 is 10 and 80% of 84, 80 is 64. So you got how much? 74 out of 100. Yeah, if you would have got 50 out of 100 in internal and 80 out of 100 in the final paper, so you will be getting 74 out of 100 in the final score. So this is weight. So what did I do? I multiplied 0.2 into 50 your score plus then 0.8 into the second score and then the total score I took. Okay, so this is how, yeah, so this is what you will be getting in the final exam. So this is the concept of weightage. So now let us say there are A1, A2, AN numbers and W1, W2, W3 and WN are weightages. Then if I define A as, let us say A as A1, W1 plus A2, W2 plus A3, W3 like that, so on and so forth. And this is AN, WM, right? This is weighted arithmetic mean. This is called weighted arithmetic mean. It will not be N, it will be A1 plus, sorry, it will be W1 plus W2 plus W3. Plus W4 plus dot, dot, dot, WN. So if you see here, here that we have to divide by 20 plus 80, weights were or 0.2 by 0.8, 0.2, 0.2 plus 0.8. So that's what you have to divide. So here it will be 0.2 and 0.8. These were the weights here, right? So this is A and then let us define G and what is G? G is nothing but A1 to the power W1 times A2 to the power W2 times so on and so forth, AN to the power WN whole to the power 1 upon, 1 upon W1 plus W2 and all that. So I will write like this. So this is 1 upon power is 1 upon W1 plus W2 plus dot, dot, WN. If this is G and H is defined as W1 plus W2 plus dot, dot, WN divided by W1 upon A1 plus W2 upon A2 plus dot, dot, WN upon AN. These are three A, G and H I found out. So in this case also A is greater than equal to G is greater than equal to H. This is weighted means. This is the concept of weighted means. So let me give you a few more theorems before we actually solve some more problems on that. So now let me give you this. What is power mean equality? This is what did we learn? This is weighted means inequality. So this is one and then now it is something called power mean inequality. So power, power, mean, power mean inequality. So could you scroll up? Scroll up. You are writing now is it? Yes, sir. Thank you, sir. Now power mean inequality. What is it? So again, we are dealing with positive numbers A1, A2, AN. All belongs to R plus and W1, W2, WN. All again are non-zero real numbers. Sorry, all are again are positive real numbers and let us say M be a non-zero real number. Then power mean WPM for M is given as nothing but weighted mean of all. So let us say this is W1, A1 to the power M plus W2, A2 to the power M plus 1 and so forth plus WN, AN to the power M divided by all the weights W1 plus W2 dot, dot, WN and this whole 1 by M. This is defined as WPM, power mean of any N numbers like that. So hence, let us say there are two numbers P and Q. So that's that P is greater than Q. Then WPM of P will be greater than WPM of Q. This is another inequality and I will take an example and show you. So is the bracket for the numerator only or the numerator can denominator? Let us take an example and see the application of power mean inequality. So what is the application? Application is prove that A to the power 4 plus B to the power 4 plus C to the power 4 is greater than equal to A, B, C, A plus B plus C. And it is given that A, B and C all belongs to positive real number set. So how to prove that? So let us now try to find out weighted power. So weighted mean WPM 4. What will be WPM 4? So WPM 4 will be when weight is let us say 1. All the weights are 1. So hence you can say WPM 4 is A. Let us say all the weights are 1. So how do I formulate this? So let me first write. So let us say A, B, C are three numbers all greater than equal to 0. Then let us say W1, W2, W3 the weights or in this case not 1, 2, 3 in case in this case the weights are A, B and C. So weight of A is let us say WA. Weight of B is let us say WB and weight of C is WC all are equal to 1. So now see here. So 1 is the, so in this case if I just now put all these values. So let us say WPM 4 will be nothing but 1 times A to the power 4 plus 1 times B to the power 4 plus 1 times C to the power 4 divided by 1 plus 1 plus 1 and whole to the power 1 upon 4. This is WPM 4. Correct. Now what will be WPM 1? WPM 1 will be 1 times A to the power 1 plus 1 times B to the power 1 plus 1 times C to the power 1 divided by 1 plus 1 plus 1 whole to the power 1 upon 1. Isn't it? Now by the inequality which we just learned WPM 4, if 4 is greater than 1 then WPM 4 will be greater than WPM 1. In fact, greater than equal to if, if guys just excuse me for a minute. I am getting some, getting some urgent call. Yeah, sorry guys. So hence we were here. Yes. So hence WPM is greater than equal to, so the inequality is, so P is greater than equal to, this is equal to equal to, so WPM is greater than equal to WPM 1. Okay. So you are not getting the just a minute. Got it. What happened? Just a minute guys. Do some arrangements. Just, sorry for the inconvenience. So this is what yeah. So WPM is greater than equal to WPM 1. This is what we learned and now we will be applying. So hence if you apply this what do you get? You get what? So let me write that. So A to the power 4 plus B to the power 4 plus C to the power 4 divided by 3 whole to the power 1 upon 4 will be greater than equal to A plus B plus C by 3. Isn't it? A plus B by C by 3. Now if you raise both sides to the power 4, you will get A4 B4 plus C4 divided by 3 is greater than equal to A plus B plus C by 3 whole to the power 4. Is it correct? So that means now this can be simplified as so what was our target? So we need one A B C and A plus B plus C. So let A plus B 1 A plus B plus C count of it. So hence you can say this is the right hand side or the left right hand side is A plus B plus C by 3 times A plus B plus C by 3 to the power 3. Okay. Right. Now A plus B plus C A plus B plus C A plus B plus C by 3. What is it? It is giving me an indication of AM of A, B and C that will be greater than equal to A, B, C to the power 1 by 3. Yes or no? Right. So hence if you raise both sides A plus B by 3 by 3 to the power 3 will be greater than equal to A, B, C. Yes. So hence what do I get? I get on the this particular side will become what? This will become this can this is now greater than equal to A plus B plus C by 3 and this is A plus B plus C by 3 whole to the power 3 is greater than A, B, C. So I can multiply this by A, B, C. Isn't it? So that means this is greater than A plus B by so hence yeah. So hence what do we what do I get at the end of the day? I get here this one we started from. So hence it is A to the power 4 plus B to the power 4 plus C to the power 4 by 3 is greater than equal to A plus B plus C by 3 times A, B, C. So this three and this three case cancelled so hence we get the desired inequality. So you can try other methods also other methods also can fetch you the same but this is an application of weighted what is that power mean inequality. So this is okay moving ahead then there are some more important inequalities and one is and now the one which we are going to discuss is also very important and very famous and this is called Chef by Chef's inequality. Okay this was a Russian guy who gave this inequality in the score Chef Chef Chef by Chef's Chef by Chef's inequality. So last time we saw we stress inequality now this is Chef by Chef's inequality and it says let xi yi belongs to r okay for every for every i is equal to 1, 2 till n what does it mean? It means simply that x1, x2, x3 so i is equal to 1, 2, 3 like that xn and y1 y this is another way of writing this y2, y3 dot dot dot yn these are all real numbers such that x1 is greater than less than equal to x2 is less than equal to x3 less than equal to dot dot dot xn and y1 is less than equal to y2 is less than equal to y3 is less than equal to dot dot dot yn then then what is this inequality x1 yn plus x2 yn minus 1 plus dot dot dot plus x3 y sorry not x last will be xn so this is xn y1 right this whole divided by n is less than equal to x1 plus x2 plus dot dot dot plus xn whole divided by n that is arithmetic mean of xi's and similarly y1 plus y2 plus dot dot yn divided by n so arithmetic mean of y and this whole is less than equal to x1 y1 plus x2 y2 plus x3 y3 plus dot dot dot xn yn divided by n okay this is what is called Chebyshev's inequality okay now sir yep for the second part of the inequality is it x1 x plus x2 plus xn all that into y1 plus y2 plus yn by n yes okay this one is x1 summation xi divided by n this is summation yi divided by n multiplication okay so now let us to understand this is Chebyshev's inequality now keep this inequality in front of you and try to solve this one okay if a b and c belongs to r plus all are positive this is question on basis this then prove that a to the power 8 plus b to the power 8 plus c to the power 8 divided by a cube b cube c cube is greater than equal to 1 by a plus 1 by b plus 1 by c prove it prove it so you'll have to apply Chebyshev's just a minute guys any luck yes sir what happened no clue no clue okay I'll give you a hint but what does Chebyshev say okay so you have to have two xi's and you know yi's first of all isn't it so if you take the first half of this this is very vital so let us say if I take xi is equal to a to the power 6 and yi is equal to let us say a to the power 2 like that xi yi and similarly so x1 you can consider so x1 is equal to a to the power 6 and y1 is equal to let us say a square similarly x2 you can consider b to the power 6 and y2 is equal to let's say b square and then third is x3 is equal to let us say c to the power 6 and y3 is equal to c square now check this this part of the Chebyshev's the Chebyshev's right hand side the two x3 right so sorry the second and third term what does it mean it says let us say if n is equal to 3 so x1 y1 plus x2 y2 plus x3 y3 divided by 3 okay this is greater than equal to what x1 plus x2 plus x3 by 3 whole into y1 plus y2 plus y3 divided by 3 is it it yes so this is what we would have gotten so hence if you see now on the left hand side x1 into y1 will be nothing but a to the power 8 and then second term is b to the power sorry b to the power 8 yes and third term is nothing but c to the power 8 divided by 3 divided by 3 and that is equal to greater than equal to that is greater than equal to x1 plus x2 plus x3 so what do what do I get here here we get a to the power 6 plus b to the power 6 plus c to the power 6 right and this is this divided by 3 okay and then multiplied by this a square plus b square plus c square divided by 3 isn't it that means what do I get I can say that 1 3 is cancelled 1 3 will be cancelled so this is cancelled this 3 will be cancelled so eventually the inequality becomes what inequality becomes 3 times a to the power 8 plus b to the power 8 plus c to the power 8 is greater than equal to is greater than equal to a to the power 6 plus b to the power 6 plus c to the power 6 times a square plus b square plus c square okay this is what the first first part of the inequality is somewhat I now we are getting a to the power 8 plus b to the power 8 plus c to power 8 now let us see whether we can go further now a to the power 6 plus b to the power 6 plus c to the power 6 guys okay by amgm inequality what can we say this by 3 is greater than equal to a 6 b 6 c 6 whole to the power 1 by 3 which is nothing but a square b square c square okay similarly similarly what can we say is there is one one more step so this is one thing which I didn't teach you but anyways please remember this that a square plus b square plus c square right a square plus b square plus c square will be greater than a b plus bc plus ca why why will it be like that because if you see um yes yes if you multiply both sides by 2 so it will be like that so hence you can see what you can say that a square plus b square minus 2 a b then plus b square plus c square minus 2 bc plus c square plus a square minus 2 ac this whole should be greater than equal to zero why all of them are square so if you see a minus b whole square plus b minus c whole square plus c minus a whole square is greater than zero so from here it follows this right so what can I what do I learn I learn a square plus b square plus c square is greater than a b plus bc plus c a greater than equal to that correct now a square plus b square is greater than a b plus bc plus ca and and what else I got um yeah and so there are two parts of this inequality here this part so if you if you take this three here so it becomes three a square b square plus c square so you can cancel this three right so that means a to power six b to power six c to power six all added together is greater than this and a square plus b square plus c square is greater than a b plus bc plus ca that means this whole left hand side can be written as three times a to the power eight plus b to power eight plus c to power eight is greater than equal to what is greater than three times a square b square c square times a b plus bc plus ca so hence if you see this three this three goes so what is left so this is what so hence I get um what do I get I get uh yeah so this is so from here I have to get a b plus bc plus c okay so now what what is the uh original inequality original inequality says divide it by a cube b cube c cube the a eight b eight c eight so hence dividing both sides by a cube b cube c cube we have to divide a cube b cube c cube so hence the right hand side inequality is one by a bc divided by a b plus bc plus ca hence it is one by what one by c plus one by a plus one by b isn't it so the right hand side we arrived at whatever was required c one by a plus one by b plus one by c so hence this inequality is through did you understand this guys yeah did you understand how to apply uh shea vices and multiple other uh together you know you know multiple other things yes so so this is please keep in mind and uh this is how you have to you know obviously you have to solve a lot of problem spaces them then only you will be clear so I'll be sending you some problems and then you can apply them now another very famous inequality is Cauchy Schwartz inequality okay so what is that inequality I'm writing it here Cauchy Schwartz Schwartz inequality Cauchy Schwartz inequality again so this guy was a french guy and this guy was again a german so this guy was french Cauchy Schwartz was a german prussian guy and in early 18th sorry 19th and 20th century now what is this inequality again if a1 a2 a3 dot dot dot an and b1 b2 b3 and bn right are real numbers real numbers then then then then what a1 b1 plus a2 b2 plus dot dot dot plus an bn an bn squared squared is less than equal to less than equal to a1 squared plus a2 squared plus dot dot an squared into b1 squared plus b2 squared plus dot dot bn squared okay and equality holds equality holds holds where does the equality holds equality holds only if if a1 by b1 only if a1 by b1 is equal to a2 by b2 is equal to c2 by sorry a3 by b3 is equal to all that an by bn this is called Cauchy Schwartz inequality so if you see the previous one what was the any any resemblance some closeness so Chewbyshev's inequality was involving arithmetic mean and all that you know there was no powers here right so hence arithmetic mean of two series all must be n all n numbers were there right so x1 x2 x3 xn to be by n then y1 plus y2 plus yn all this here we are talking about in this particular step these are am and two am's of two series of number is less than if you see one product take two of two taken taken at a time so x1 y1 corresponding elements if you multiply and then take the mean so hence the multiplication mean will be more than the product of individual mean this is interesting and this is on the on the left hand side it is the reverse order of multiplication so this is there's no power here now here they talk about not means here it is power right so again multiplication of a1 b1 a2 b2 and all that whole square is less than equal to the so sum of the square of the sum basically you can say the square of the sum is less than is less than the what product of some of the squares okay this is what in another way you can say now can you prove it can can anyone try and prove this how can this be proven if you use vectors then it's the dot product if you what is vector if you use a vector in and in multiply them i think you get this inequality i i'll do it and let you know yeah try try try to prove this try and tell me how can this be proven and then i'll give you the proof yeah yes i got it so yeah tell me so if you take the two vectors u and v where u is a a1 a2 a3 until a n and v is b1 b2 b3 until bn the dot product of the of the vectors is the magnitude of u into the magnitude of v into cos theta so that is um so the that is less than equal to the magnitude of u into the magnitude of v like so u v cos theta is less than u v yeah what you are trying to say is this that i don't know if most of others are comfortable with vectors but let us see let us say vector x is given by let us say a1 and it is n dimensional vector let us say so a1 and or in terms of matrices also you can talk about but let us say yeah but most of you would not be comfortable with the dot product things and let me not do it here i'll i'll give you some other this thing but you are right sarin so the the magnitude of uh you know the magnitude of a vector and the product of magnitude of two vector will be greater than equal to the dot product of the vectors yeah so that one way of that during that is this otherwise what you can do is this that yes so again we will be using quadratic uh you know inequality right so what will what will we do so we are going to define a function is like this so let us say fx is equal to fx is equal to i am writing a1x minus b1 whole square plus a2x minus b2 whole square plus dot dot dot plus anx minus bn whole square okay and do you think this is greater than equal to zero because all are sum of squares isn't it so if you open it up what will you get you'll get x square common and within brackets you'll get a1 square plus a2 square plus dot dot an square and then you'll get minus two x common and within brackets you can write a1 b1 plus a2 b2 plus dot dot dot an vn and then final term is nothing but what b1 square plus b2 square plus dot dot dot bn square correct and this is greater than equal to zero so that means the discriminant of this must also be greater than equal to zero sorry less than equal to zero so hence what is the discriminant here so discriminant clearly is d is equal to a1 b1 plus a2 b2 plus dot dot dot an bn this is current b square so this is square minus four times ac so ac will be a1 square plus a2 square plus dot dot dot an square times b1 square sorry sorry thank you forgot coefficient minus two this is four right so b square and then thanks for correcting and then this is bn square right so this is my uh what d and this must be less than equal to zero so hence it automatically proves the theorem what is the theorem now so hence a1 b1 plus a2 b2 plus dot dot dot an bn whole square is less than equal to a1 square plus a2 square plus dot dot dot an square times b1 square plus b2 square plus dot dot dot bn square okay so this proves the theorem now okay so let me give you a problem on this then hmm which one should i give um a apply b plus b by c somebody a square this is some a little bit problem okay let's try this so question is question is this if p1 p2 p2014 so you can guess this is a question of 2014 bn arbitrary arrangement okay so these are um they are 2014 real numbers okay proves the inequality 1 by p1 plus p2 plus 1 by p2 plus p3 plus 1 by p3 plus p4 so on and so forth plus 1 upon so this is a perfect rmo question 2013 so this is a perfect rmo question p2013 plus p2014 and this is greater than equal to 2013 by 2016 so your hint is you have to apply cautious shots inequality the inequality is this just above guys one thing is uh see all these are 2014 real numbers but all of them are from 1 2 3 2 2 0 1 4 okay understood so p1 p2 p3 p4 all are either 1 2 3 or something like that so there are 2014 numbers but which one is what is not known but they are certainly 1 2 3 till 2014 so use that concept yeah yes struggling yes guys any idea yes sir did you get anything how many numbers are did you understand the question the question is something like this the question is 1 2 3 4 till 2014 1 2 3 4 till 2014 these are the numbers okay and and all these numbers are any so let's say this one if this p4 is p1 then this is p2 then this is p3 like that so there are 2014 such numbers ranging from 1 to 2014 now you have to find out you have to prove this inequality so and i told you that cautious shots can be applied so see here p1 plus p2 what i'm saying is p1 plus p2 p1 plus p2 okay let us observe these numbers and p2 plus p3 and p3 plus p4 all of them these are the numbers here can you see let these numbers be let these numbers be let us say this is my a uh this is my b this is c okay like that uh wait uh yeah so instead of writing a1 b1 so i'll be writing let us say this is a1 this is a1 this is a2 sir i got it yeah how but i didn't use the method we were supposed to actually let me just call this method i'll because the idea is to explain you the use of this inequality so let us say this is all these are all the term so what is the final term final term will be a201 sorry p2013 plus p2014 so this is term number how much a2013 isn't it now consider other terms like one upon p1 plus p2 and call this one as b1 okay and then one upon p2 plus p3 and call this as b2 and so on and so forth bn will be b what b2013 will be nothing but one upon b2013 plus b2014 isn't it so uh this is p not b i am now now what koshy short says that a1 b1 plus a2 b2 plus a3 b3 all dot dot dot an bn right squared is less than equal to what a1 squared plus b1 squared plus b sorry a1 squared plus a2 squared plus a3 squared plus dot dot dot an squared right multiplied by b1 squared plus b2 squared plus b3 squared plus dot dot dot bn squared bn squared isn't it am i right or i'm sorry uh yes this is an equality right so hence to beat it what i can do is um okay a1 squared it is right now so so so so what can i do is a1 squared plus a2 squared plus an squared okay now instead of that what other thing which i can do is i can i can say let this be roots be the terms all roots okay why am i doing this you will get to know quickly so let us say the roots are there then if you multiply the a1 and a1 and b1 what will you get you will get one isn't it similarly if you multiply a2 b2 you'll get one you get a3 b3 you'll get one so hence i can say in the left hand side it will be one plus one plus one plus dot dot dot how many n is nothing but two zero one three so it is one two two thousand thirteen ones are there so this square okay and then on the left hand side a1 square is what p1 plus p2 is it and a2 square is p2 plus p3 is it and likewise an square is p2013 plus p2014 and this multiplied by what one upon b1 square is p1 plus p2 and then plus one upon p2 plus p3 plus dot dot dot it is one upon p2013 plus p2014 is it okay right so on the on the this thing you'll get 2013 squared right 2013 squared is less than equal to I am writing p1 plus p2 times sorry plus p2 plus p3 plus dot dot dot p2013 plus p2014 multiplied by one upon p1 plus p2 plus p1 upon p2 plus p3 plus dot dot dot one upon p2013 plus p2014 isn't it now if you closely look at it what can I say about this so hence I can say 2013 whole square divided by let me take this item on the left hand side so what is the left hand side thinking so this will be 2013 square divided by p1 plus p2 plus p2 plus p3 so on and so forth plus dot dot dot p2013 plus p2014 okay and this would be less than equal to one upon one upon what is it one upon p1 plus p2 plus dot dot dot one upon p2013 plus p2014 guys am I right so far so good so hence if you see the what is the denominator of the left hand side if you closely see this is nothing but 2013 square now I'm only dealing with LHS guys okay so LHS 2013 square and this is nothing but every item is two times but p1 and p2014 isn't it so hence can I not write this as two times p1 plus p2 plus dot dot dot p2014 minus p1 minus p2014 isn't it yep because every every item was two times so hence I can write this right now what is p1 p2 p3 basically it is an ap isn't it one two three four till 2014 is it so what is it nothing but 2013 squared divided by two times sum of an ap is number of terms 2014 divided by two into number of terms plus one is it minus p1 minus p2014 okay isn't it so that means this is nothing but this to this to go so it is equal to 2013 squared by 2014 into 2015 minus p1 minus p2014 so far so good now if you see this item p1 and p2014 right so what is if I increase the denominator then but then the value goes down and if I decrease the denominator then the value goes up so this particular value is this particular this particular value is greater than equal to I am saying 2013 square divided by 2014 into 2015 minus one and minus two why is this so because one and two are the least of the all 2014 numbers right so this this particular so here if you see the numerator is same but the denominator on the left hand side is lesser than denominator on the right side is it okay right yes so yes so what can I now say so hence now okay so then what okay so this must be greater than equal to this must be now again I can simplify this and what can I say about one or rather what I could have done is yeah so I can remove one more so this must be greater than equal to 2013 square 2014 times 2015 minus one minus one so I am reducing the denominator further is it did I do this correctly am I I am reducing the denominator no I am increasing the denominator actually I added one in the denominator so hence the right hand side will go down is it it the denominator is increasing now this this item can be written as 2013 square divided by 2014 2015 and minus one can be written as 2014 minus 2015 and minus one can be left as it is now this is on purpose why this is now 2013 square divided by I can take 2014 common from what from yes so this is 2014 common and I can write this particular expression here can be written as 2015 minus one and 2015 plus one am I right yes this is a factorization isn't it so it is basically nothing but 2013 by 2016 which I got the answer as so to basically this is less than the given problem statement and this is what we have to prove isn't it what was it oh sorry so we had to prove that this is where did I start yeah so the the 201 through by to the 2016 is less than all that activity which we did so this is this is all the other terms are more than this so hence proved okay so you have to do a lot of mental exercise to get to the solution it's not that simple like you know other your syllabus and the board syllabus that you apply the formula and you get it there to think through think through and you know I have to work on lot many steps to eventually find out the solution and the only way out is not me solving the problems but you you guys solving the problems yes keep those tools in mind so that whenever you require them you can use them but ultimately you know your thinking process only will decide whether you're gonna able to solve the problem or not okay so I hope we are good we are done with any inequalities this much knowledge is good enough to clear RMO I think in inequalities later on now from next class we'll be having some other topic okay so I'll be sending some problems for you please solve and let's see you know and those inequalities now try and solve those questions and you must solve at least one question on a daily basis otherwise there is very you know it will be very difficult to uh catch the pace okay fair enough guys so let's call it a day all the best bye-bye bye sir bye bye so bye