 Hi, it's Zor. Welcome to Unizor Education. Today we will continue talking about mass plus and problems, which is part of the Unizor.com. Now this course mass plus and problems contains different problems, which are not really standard, just to check your theoretical knowledge. They are a little bit about thinking. Non-standard, if you wish, non-typical. So today we will talk about averages. I do recommend you to watch this lecture from the Unizor.com because every lecture, including this one, contains textual part, notes, basically, which is like a textbook. And in general, all the lectures are logically related. There are menus, so the site is, by the way, absolutely free. There are prerequisite courses here in this particular case. The prerequisite course would be mass for teens. There's also physics for teens on the same website. All totally free, no ads, no signing required. I mean, if you wish you can. So just very good source of information. Okay, so, talking about averages. Well, first of all, there are many different types of averages. For example, everybody knows something like a arithmetic average or mean. Let's say you have number four and number nine. So you basically add them together and divide it by two, and that would be 13, like 6.5. That's a arithmetic average. There is a geometric average, which is the square root of their product, which is what, 366. Well, 6.5 is in between. Six is also in between, so it's different averages. There is a quadratic average, which means you have to put four square plus nine square divided by two and square root. That's what, 16, 81, 97 divided by two, 48 and a half, almost 49, so it's approximately seven. Also in between, but it's different averages. Now, what's interesting is that arithmetic average is always greater or equal to geometric average. And that's what we are going to prove today. And not only for two numbers, but for any number of numbers. So, now, here is a set of numbers where index i is changing from one to n. Now, a arithmetic average is, you have to summarize them, and divide by number n. That's a arithmetic average, and I call it a, and a stands for arithmetic. Now, geometric average, let's use the letter G. That's, you have to multiply them, all of them. So, it's a product. It's a Greek letter pi, also from one to n, x i. And have the root of n's degree, which is one over n power, as you know. Okay, so what we have to really prove is this. Okay. Now, there are many different proofs. It's not really like an easy thing, by the way. So, there are many different proofs. I will present two. And one of them, the first one which I'm presenting is by a famous mathematician French, I believe, Cauchy. Okay. It's almost like by induction, but not exactly. So, first, we check it, obviously. That's the approach which we're using all this to check for induction, to prove by induction. So, for n is equal to 1, that's obviously equal to, it's only one number. So, it's, some would be that number itself, x first, divided by 1. So, it's still x 1, multiply x 1 from i is equal to 1 to 1, basically. So, it's also x 1. And the square root of the first degree, which is also x 1. So, it's x 1 and this is x 1. So, it's equal. So, that's equal. Now, with 2, it's just a little bit more difficult. But still, not very difficult to prove. With 2, I will do something which is, I believe, general kind of a logical approach. Here it is. So, you have, let's say, two numbers, x 1 and x 2. So, their arithmetic average is this. Now, their geometric average is this. I use square root. Now, I didn't mention it before, but we are talking only about non-negative numbers. Because otherwise, square root would have some problems, etc. So, it's real non-negative numbers. Now, I would like to prove that this is greater than this. Regardless of what kind of non-negative real numbers x 1 and x 2 are. Now, I will prove it the following way. I will start from something which we have to prove. This is some kind of analysis stage. I will make certain transformations which are invariant and reversible. And obviously, what's irreversible? Well, for pluses minus, for square root, it's power. And for division, it's multiplication, vice versa. So, in many cases, these are reversible transformations. For example, if we don't restrict ourselves to non-negative numbers. If we allow negative numbers, then square root and power of 2 are not really reversible. Because from minus 5, you can raise it to the second power of 2, that would be 25. But if you will extract square root, arithmetic square root, you will have 5, not minus 5. But within the non-negative numbers, that's a reversible transformation. So, I will use reversible transformations to come up with absolutely true statement. Is it a proof? No. Because from a false, we can always get to true using logical transformations. So, the proof is from absolutely true statement, logically come up with whatever we need. So, in this case, I will just reverse all the transformations. And that's the proper way of proving certain things. So, how do I do it? Well, I will square it, the both parts. So, I will have x1 plus x2 square divided by 4. And I will try to prove it that this is x1 times x2. Or x1 square plus 2x1x2 plus x2 square greater than 4x1x2. 4 goes to the left part. I will have x1 square minus 2x1x2 plus x2 square greater than 0. Now, this is obviously x1 minus x2 square greater than 0. Now, this is absolutely true statement. x1 minus 2, since it's square, is greater or equal to 0. Now, I'm saying all the transformations which I was just making are reversible invariant transformations. From here, from the absolutely true statement, I can go to here. From here, I can go to here by adding 4x1x2 to both sides of equation. That's an invariant transformation. Now, this I will transform into this. Just, you know, change the explicit square into this and divide it by 4. And from this, by extracting square root from both sides, which is an invariant reversible transformation if we are talking about non-negative numbers, I get this. Now, this is the proof. So, analysis, you go from whatever needs to be proven to definitely a true statement. And then the real proof is to go backwards. And it's not obviously in these only cases, it's many other cases. That's the technique which you have to make, which you have to go through if you want to really prove the thing. Some people, for whatever reason, just stop at the very end of this. Start from this and come to this and say, okay, that's a proof. No, that's not a proof. Because it might be the false statement and you still can come up with a true statement. We did these examples, man. We were talking about logics in the mass-vegetance course, prerequisite course. Okay, so, for one, we have checked, for two, we have proven. For n equals 2. n equals 1. We check n equals 2, we prove. The proof is not exactly the way how we do it in, I would say, typical inversion. Induction, sorry, induction. We were usually going from n to n plus 1. So, from 2, we can go to 3, from 3 to 4, et cetera, et cetera. Now, we will do it differently. We will do it from n to 2n. So, I assume that for n, it's true. An greater than gn. Arithmetic average of n numbers is greater than equal to a geometric average. So, I assume that for n, it's true. Then I will prove that for 2n is also true. What it will give me? It will give me from 2, I can go to 4. From 4 to 8, from 8 to 16, from 16 to 32, et cetera. Only powers of 2, only if n is powers of 2, that would be the proof. Okay? So, that's not a complete proof. So, that would be a continuation. But the first thing I would like to prove is this one. So, again, my first stage, I check and prove for this. My second stage is this. And the third stage would be for all other n's. Okay? Now, how to prove it for this? Okay. So, if you have 2n numbers, let's divide it by two subsets. Okay? Subset number one would be xi where i is from 1 to n. And subset number two would be all xi's where i belongs to from n plus 1 to 2n. So, I have divided my group of 2n numbers into two groups. From 1 to n and from n plus 1 to 2, 2 to n. Now, each one of them has n elements. So, for each one of them, this is true. So, a1n greater than g1. Now, one means for the first subset. Now, for the second subset, I have a2n greater than g2n. Okay? Now, what is a2n, which is arithmetic average of all of them? What is this? Well, that's sum of these and sum of these. And then you sum them together and divide by 2n, right? So, what is the sum of these? Well, if I know the average, if I average multiply by n, that will be sum of these, right? n times a n first. Now, sum of these are, they're also number n times a n 2. And then I divide it by 2n, right? So, this is a n 1 plus a n 2 divided by 2. Okay. Now, what is their geometric average? gn, no, g2n. g2n is their product in power 1 over 2n, right? So, what is their product? Well, product of the first one, since we know geometric average, that would be geometric average of the first one. And I will raise it to the power of n, right? So, this is product in the power of 1 over n. But if I will multiply, if I will raise it to the power of n, I will get the product itself. Then I have to multiply it by another of the second group, power of n. And all together, I have to raise to the power 1 over 2n. So, this is a product of the first n of the first group. This is the product of the second to the power of n. And then, from all the product of all the numbers, I have to raise to power 1 over 2n, or root of the 2n's degree of power. Okay, which is equal to what? It's equal to gn in the power of my rate, times gn, the second one, in the power of 1 over 2. Powers are multiplied, right? Which is gn times gn to the power of 1 over 2. Now, okay, that's done. So, I have my expression for g2n, and I have my expression for a2n. Now, the rest I can wipe out. So, this is sufficient. Now, a2n is equal to an first plus an second divided by 2. So, all the 2n's is basically average of two different groups. Now, this is greater than or equal to, I assume that for n, my inequality is true. So, this is greater than gn first plus gn second divided by 2. I just replace this with this, and this with this. That's why I'm putting the greater than sign based on this. Now, these are two numbers divided by 2, which is arithmetic average of two numbers. It's greater than geometric average of these two numbers, which is what? gn first times gn second square root. But this is exactly this, equals to g2n. And that's how we have proven that a2n is greater than g, or equal to g2n. So, arithmetic average of 2n numbers is greater or equal to geometric average of these two numbers, but under assumption that for n, it's true. So, now, since we know that for n equals 2, it's already true we have proven it. So, from n equals 2, we go to n equals 4, from 4 to 8, from 8 to 16, to 32, etc. So, we have proven for all numbers n. So, for all n is equal to 2 to the power of k, where k is some natural number, we have proven it. Now, how to do in between? So, let's say my n is between 2 to the power of k1 and 2 to the power of k. So, it's somewhere in between. So, you have this xi, where i from 1 to n, and n is here. So, what I will do is, I will complement this set of n numbers to a set of, let's say, m is equal to 2 to the power of k minus n. This is number of elements which I need to complement to 2k. So, I will put, let's say, g n m times. So, my new set contains n given elements, and m, where m is this, g n, where g n is geometric average of these n numbers. So, all together I have n plus m, which is 2 to the power of k numbers in my set. So, x1, x2, etc., xn, and then g n, g n, g n, g m times, all together constitute a set of 2k to the power of k numbers, for which my previous proof has been already done. So, for 2 to the power of k, I know that everything is true, which means that x1 plus x2 plus, etc., plus xn, that's my sum, plus g m, and plus g n, plus, etc., g n. So, these are n elements, and these are m elements. I know that their arithmetic average, which is divided by 2, is greater than equal their geometric, which is what? Product of all of them, which is product for x i, i from 1 to n, times product of all my g n, the m, the m of them, so it's the power of m, and then I have to have power 1 over m plus n, right? m plus n. So, let me just write it again in a more compact form. So, this I have to divide by, well, I don't know why I put 2 here, I have to divide it by m plus n, right? All right. So, what do I have now? I have sum of x i, i from 1 to n, plus m times g n. This is the left part, and all of this is divided by 1 over m plus n, right? So, that's my left part, and I know this is greater than or equal to product from 1 to n of x i, times g n to the power of m, and power 1 over m plus n. Okay, that's good. So, this is a n, and this, sorry, this is a m plus n, and this is g m plus n. Okay. Now, let me just simplify it. What is this sum? Well, this sum is basically the average times number of elements. So, it's a n times n. That's this one, plus g n times n. So, that's my left part, right? Divided by m plus n. On the right, I have... Now, what is this product? Product is g n to the power of n, times g n to the power of m, and all of this is power f plus 1 over f plus n. So, that's how my inequality looks now, and this is a true inequality because it contains m plus n to the power of k elements. And basically from here, I will derive whatever I need. Okay. How do I do it? Now, what is this? This is g n power of n and power of m. So, together it's m plus n, and then 1 over m plus n. So, this is just g n on the right. So, on the right, I have g n. No. m plus n and 1 over m plus n. On the left, I have this. a n times n plus g n times n. And I will put this m plus n to the right side. So, what's left? g m, g m. So, what's left would be a n times n, greater than g n times n, because g n times n will cancel out, cancel this, and I have a n greater than g n. Arithmetic average greater than... Now, I can actually do exactly the same by complimenting, not with geometric average, to complete my set to 2 to the power of k elements. I can use the arithmetic average and do exactly similar kind of things and get the similar result. I put it in notes for this lecture. So, that's probably would be a nice exercise for you when you will read it. It's basically the same kind of proof, but slightly different. And then I wanted to have another proof very quickly, which belongs to Podia. Very clever, I would say. Now, you know that this is e to the power of x graph. It has tangential line at x is equal to 0 at 45 degrees. So, the tangential line equation would be y is equal to x plus 1, obviously. So, it's x raised by 1 up. So, if I will use... If I will shift all the graphs to the right, I will have... Instead of this point, I will have this point. And the tangential line would be at point x is equal to 0, 0. Not 1, 1. Here, 1, 1. That's tangential point. 0, 1. We shift to the right. Now, when you shift to the right, the graph of the e to the power of x would be e to the power... Sorry. e to the power x minus 1. And instead of y is equal to x plus 1, I would have is equal to x. Now, I can actually spend some time talking about the following equation. Now, from the graphical standpoint, you obviously see that e to the power x minus 1 is greater or equal than x, right? This is greater than this. It's above. Now, we can actually go to some calculus thing and saying, okay, the first derivative at this point is equal to 0 between the difference of these two functions. And it never actually goes to 0 anywhere else. It's always positive. So, if I will have difference between two functions, and that means that it's convex, and that means that it's always positive. If it's 0 at one point and it goes only up. But it's kind of obvious right now just from the graphical standpoint. And they don't want to go into these calculations. So, I will take this as basically as granted and obvious inequality. But from here, it's very easy to have our inequality because I will use it for x i divided by a n. So, what will be? I will have e to the power x i divided by a n minus 1 greater or equal to x i divided by a i. So, instead of x, I'm using this. Then, I multiply it from 1 to n. What will I have? I will have e to the power of x 1 divided by a n minus 1 minus e to the power for x 2, for x 3, etc. And powers are aiding together. So, if I will add them together, I will have basically sum of x i divided by a n minus 1 would be n times. That would be on the left part. On the right part, I will have product of all x i's at 1 to n divided by a to the n to the power of n. Right? Because a n would be multiplied by each other. Now, what is this? Well, sum divided by n is a n. Sum divided by a n is n. So, n minus n would be 0. So, this is 1. That's why we have that 1 is greater than product of x i's divided by a n to the power of n. Or power of a n to the power of n greater than product of x i. And if I will use power 1 over n, I will have, on the left side, I will have a n. On the right side, I will have g n. That's the definition of the geometric average. So, from this very simple inequality, this greater than this, I can very easily derive this. That's a very clever actual idea. All right, so that's it. I suggest you to read the notes for this lecture. I put a little bit more, maybe, some logic behind these graphs, but it's obviously a simplest calculus problem, standard calculus problem, that this is greater or equal than this one. Other than that, everything was really kind of strong proof, two different proofs, that arithmetic average is greater than geometric average. And there are other proofs, if you want to. It's all brain exercise. So, I do recommend you to spend some time and get into this proof very, very close. So, you can read these notes for this lecture. To get to the notes, you have to go to unizord.com, choose the course mass plus and problems. This is algebra and this is algebra 2.0.2. That's it. Thank you very much and good luck.