 So this book is supposed to be one of the best books that we have in India for the people who are preparing for the J-Main and advancing exams But you know what? I was really disappointed to see that they have not provided the proof a Generic proof for the AMGM HM inequality They were not talking about a generic proof for the AMGM HM inequality However, they have tried to prove it for just two numbers, but I'm sure you are here to find the generic proof for this concept So without wasting much time, let's get started. So here all first we have to understand the concept of curves concave up and curves which are concave down So let me show you for your understanding two such curves one being a curve Which is concave up and the other being the curve Concave down now for the proof we can start with any of the two curves But as an example, I have started with the concave down curve So let me take you to the proof by taking one of these curves Which is concave down curve So here we are with a polygon on your screen, which is basically made out of joining the points a1 a2 a3 Dot dot dot till a n on this concave down polygon Let's say this curve the yellow curve is basically a function y is equal to f of x all right now on this particular polygon, I will choose the coordinates of a1 as a1 comma f of a1 I Choose the coordinates of a2 as a2 comma f of a2 I'll choose the coordinates of a3 as a3 comma f of a3 and So on and so forth till we reach a n whose coordinates are a n comma f of a n Now let us focus on a point here, which is your point g G here is the center of mass of this polygon Now we all know that the coordinates of g would be nothing, but We all know that the coordinate of g would be nothing, but it would be the arithmetic mean of the x coordinates of the points which are making the vertices of this polygon and The y coordinate of g would be the arithmetic mean of the y coordinates of all the points a1 a2 a3 till a n which are making the polygon right now What I'm going to do next is I'm going to Extend this point g vertically up Till it meets the curve again at a point which I'm calling here now as the point p Okay, so my new p is a point which has been the extension of g Vertically upwards till it cuts the curve at p Now what should be the coordinates of the point p let us focus on the coordinates of point p The point p will have definitely the same x coordinate as that of g So it will be a1 a2 all the way till a n Summed up divided by n And what about the y coordinate of p so now what would be the y coordinate the y coordinate would be nothing But the value of the function for the corresponding abscissa right Now it is very easy to figure out that the y coordinate of p Would be more than the y coordinate of g Obviously because the center of the gravity which is completely within the polygon is also lying underneath the curve So the y coordinate of g will definitely be lesser than the y coordinate of p But friends I can also say that for a very special situation The y coordinate of g can also be equal to the y coordinate of p Especially when your polygon is basically reduced to a point polygon That means if I choose my a1 a2 a3 till a n points in such a way that they're all Coincident you would realize that the polygon would be reduced to a point polygon and in such a scenario The center of gravity or the center of mass of such a polygon Will have the same y coordinate as the point p because they will all be coincident Hence I can say that the y coordinate of p will always be greater than equal to the y coordinate of g Why greater than equal to as I already told you I want to incorporate a special situation where all the points of the Polygon could be coincident So now in this situation what I'll do is I'll now take the coordinates y coordinate of p Which is f of a1 plus a2 all the way till a n Divided by n this would be greater than equal to the y coordinate of g Which happens to be f of a1 f of a2 all the way till f of a n whole divided by n So this inequality will reverse if the curve where concave up So this is for a curve, which is concave down right and For the curve which is concave up The situation will become exactly the reverse So this expression will always be lesser than equal to The sum of f a1 a2 all the way till f a n whole divided by n. Okay, so this situation will be for Concave up Right All right friends now I'm going to use one of the scenario. I do not need both the scenarios Maybe I will pick up concave down case to prove the general AMGM HM inequality Let us see which curve do we choose for this purpose? So friends for this purpose, I will choose the curve, which is your log function Okay, and for my benefit I have chosen log x to the base e However, let me tell you that you can choose any log with any base for this purpose So I have preferably chosen a base, which is greater than one so that I get a curve Which is concave? Downwards so we all know that we can call log of x as National log x also called as lawn x and if you recall the graph of lawn x the graph of lawn x Basically comes out to be a curve, which is concave downwards, which looks roughly like this Right so on this curve, I'm going to choose few points, which I'll be calling them as a1 comma lna1 a2 comma lna2 a3 comma lna3 and So on and so forth dot dot dot till a n comma Lnan Now friends the log function has always its argument as positive So for the log function to be real and defined function The input to the log function must always be positive And that's a necessary condition for us to apply the AMGM HM inequality So log function also takes care of that aspect So since this curve is concave downwards, I can use this Inequality that we had seen a little while ago to meet our requirement. Let us see how does This entire situation lead us to our AMGM HM inequality So here my f function is your ln function So I can say ln of a1 plus a2 all the way till a n Over n is greater than equal to ln a1 ln a2 all the way till ln a n whole divided by n Now here friends, I'll be using my log properties So the left hand side I will not disturb But on the right side, I can use the log property that log a plus log b is log a b So that gives me ln of the product of a1 till a n whole divided by n further by using the simplification ln of a1 till a n summation Divided by n will be greater than equal to ln of a1 a2 Product till a n whole raised to the power of 1 by n Now since I have taken a log function whose base is e is 2.718 ish Which is actually greater than 1 you can remove the log from both the sides without affecting the inequality sign So if I just take away the log or even say I can have taken the anti log on both the sides It will not affect the inequality symbol So you can clearly see that we have ended up getting a m greater than equal to gm But friends, this is just half the proof I also have to prove that gm is greater than equal to hm So if I'm able to prove that by the very same inequality concept, I'll be able to complete the proof So let's get started with the second half of the proof This time also, I'll be using the same log function So let me draw it again for you But this time I'll be choosing my points to be 1 by a1 comma ln 1 by a1 1 by a2 comma ln 1 by a2 and so on till 1 by a n comma ln 1 by a n Okay, since my curve is still concave downwards, I can call the same inequality to help us out here So f of summation of a1 till a n divided by n This is going to be greater than equal to sum of f of a1 a2 till f of a n whole divided by n Right, but this time my function input points are 1 by a1, so I'll write ln of 1 by a1 1 by a2 all the way till 1 by a n whole divided by n Will always be greater than equal to ln of 1 by a1 ln of 1 by a2 all the way till ln of 1 by a n Whole thing divided by n. So friends the coordinates which I'm taking right now is 1 by a1 comma ln 1 by a1 1 by a2 ln 1 by a2 and hence I have used those coordinates itself in my given inequality Now, let us follow the log properties from here. So this is going to be ln summation of 1 by a1 1 by a2 all the way till 1 by n and On the right side, I'll have negative of ln a1 Negative of ln a2 etc all the way till negative ln a n by n So let me further simplify this out. All right. So here at this step. I'm going to multiply throughout with minus 1 So multiplying by minus 1 would actually reverse the inequality So that would lead to ln of 1 by a1 till 1 by a n Lessor than equal to of course a negative sign will come on the left and this will become ln of a 1 a2 Etc till a n whole raise to the power of 1 by n So this was the same log property which we had utilized a little while ago and Now the negative sign can be incorporated in reciprocating the argument of the left-hand side log So if I reciprocate the argument, this is what I will end up seeing and now Since you are dealing with a log whose base is greater than 1 you can take an anti-log on both the sides without affecting the inequality sign So that would lead us to This expression the viewers can easily appreciate that this is as good as saying H m is lesser than equal to gm Okay, so here friends We have proved that the arithmetic mean is greater than equal to geometric mean and the geometric mean is greater than equal to the Harmonic means so this is what we wanted to prove So friends you can easily see here that how we could prove this a m gm h m inequality by using a very Interesting geometrical positioning of a polygon By the way, if you want to know the name of the initial inequality that we had utilized The name of that inequality is Genicent inequality. So yes, the arithmetic mean greater than equal to geometric mean greater than equal to harmonic mean is not a Fundamental inequality, but it gets its strength from the genicent inequality However, a m gm h m inequality can also be proved in numerous ways But I'm sure most of the ways that you will find on the YouTube are quite complicated or many a times They are very lengthy So, thank you so much for watching. Bye. Bye. Take care. Stay safe