I am at a beginner's level (graduation 1st Year) going through the topic of university level inequalities for the first time.

Read this recently:- "If $x_1,x_2,x_3 ,\dots,x_n$ are $n$ positive real numbers such that $x_1+x_2+\dots+x_n$ is a constant, then their arithmetic mean attains its lowest and their geometric mean attains its maximum value when $x_1=x_2=x_3=\dots=x_n=A=G$"

Wanted to understand the above statement. As per me I tested above with following and found it incorrect. I am sure I am misinterpreting it in some way:-

AM of say $4$ numbers:- $(2 + 5 + 10 + 15)/4 = 8$

Taking two extremes:- $(2+2+2+2)/4 = 2$ (looks minimum)

But: $(15+15+15+15)/4 = 15$ (which is higher than 8)

It is minimum compared to what and similarly wanted to understand Geometric mean is maximum (as compared to what)?

Appreciate help in understanding.

  • $\begingroup$ Welcome to MSE. Please use MathJax. $\endgroup$ – José Carlos Santos Mar 4 '18 at 15:35
  • $\begingroup$ You have the constraint that $x_1+\ldots+x_n$ is constant. $x_1+\ldots+x_n$ is different in each of your examples. $\endgroup$ – A. Goodier Mar 4 '18 at 15:35

That's an outright wrong way to describe it. They are holding the sum (and the number of terms) constant and looking at what happens to the arithmetic and geometric means under that constraint. In this case the arithmetic mean does not depend on what the terms are, it is just a fixed number under the constraint, so it makes no sense to say it is minimized.

The geometric mean is indeed maximized when all the terms are the same, again subject to a constraint on the sum (and a positivity requirement).

  • $\begingroup$ I think I am getting your point. It is only under the constraint and for the particular constant. Is there any geometrical representation of AM and GM to help me visualise this. $\endgroup$ – Rohan Frederick Mar 4 '18 at 16:51
  • $\begingroup$ @RohanFrederick The AM-GM inequality with $n=2$ says that a square is the highest area rectangle with a given perimeter. For general $n$ it says that the $n$-cube is the highest volume $n$-box with a given sum of edge lengths. $\endgroup$ – Ian Mar 4 '18 at 17:21

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