# why is Arithmetic mean minimum when all terms are equal

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.

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

• @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. – Ian Mar 4 '18 at 17:21