$AM-GM$ Inequality Proof From a Book

Question

Here is one of two proofs of $AM-GM$ inequality from book: "INEQUALITIES, cuadernos de olimpiadas de matematicas" by Radmila Bulajich, Jose Antonio, Rogelio Valdez:

(Improvised)

Let $A = \frac{a_{1}+...+a_{n}}{n}$. If $a_{1},a_{2},..,a_{n}$ are all equal, then we are done. But notice there will be at least two numbers such that $a_{i}<A$ and $a_{j}>A$. Because if all $a_{i}$s $>A$ or all $a_{i}$s $<A$ then we will get a contradiction.

Consider $n=4$. $A=(a_{1}+a_{2}+a_{3}+a_{4})/4$. Take two numbers, one less than $A$ and other one greater than $A$. Le this be $a_{1} = A-h$, $a_{2}= A+k$, with $h,k>0$. Notice that $a_{1}' = A$ and $a_{2}'=A+k-h$ will make $a_{1}+a_{2}=a_{1}'+a_{2}'$ but the product $a_{1}'a_{2}' > a_{1}a_{2}$.

$$ A = \frac{a_{1}+a_{2}+a_{3}+a_{4}}{4} = \frac{a_{1}'+a_{2}'+a_{3}+a_{4}}{4}$$

and $a_{1}'a_{2}'a_{3}a_{4} > a_{1}a_{2}a_{3}a_{4}$.

We can always repeat the same process and still create a number equal to $A$, and this process cannot be used more than $4$ times.

Same argument for $n$ numbers. We can always repeat the same process and still create a number equal to $A$, and this process cannot be used more than $n$ times.

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Why is the above proves $AM-GM$? I don't quite understand the connection.

Answer 1

What the proof is doing is creating a sequence of sets of $\ n\ $ numbers \begin{align} \left\{a_{11}, a_{12},\right.&\left.\dots,a_{1n}\right\}\\ \left\{a_{21}, a_{22},\right.&\left.\dots,a_{2n}\right\}\\ &\vdots\\ \left\{a_{r1}, a_{r2},\right.&\left.\dots,a_{rn}\right\} \end{align} with the following properties:

• $$ a_{1j}=a_j\ \ \text{ for }\ j=1,2,\dots, n$$
• $$ a_{rj}=A \ \ \text{ for }\ j=1,2,\dots, n $$
• $$\frac{a_{i1}+ a_{i2}+\dots+a_{in}}{n}=A \ \ \text{ for }\ i=1,2,\dots, r $$
• $$\\ \hspace{-1em} \root{n}\of{a_{i+1,1} a_{i+1,2}\dots a_{i+1,n}}> \root{n}\of{a_{i1} a_{i2}\dots a_{in}}\ \ \text{ for }\ i=1,2,\dots, r-1\ .$$

It follows from these properties, that $$ \root{n}\of{a_1a_2\dots a_n}= \root{n}\of{a_{11}a_{12}\dots a_{1n}}> \root{n}\of{a_{r1} a_{r2}\dots a_{rn}}=A\ , $$ —that is, when $\ a_1, a_2, \dots,a_n\ $ are not all equal, their geometric mean is strictly larger than their arithmetic mean.

Answer 2

It's a well ordering principal/induction proof.

I'll explain. But before I do I'll point out that I am going to be doing a lot of statements like "one of the values of $a_i > A$ so well assume that that one is $a_1$". As both addition and multiplication are commutative, I'm going to simply index and reindex the variable as I go along and assume that somehow by magic the variables were all lined up in the order I needed them in from the beginning.

So we have $a_1,........, a_n$. The average value of each of these is $A=\frac {a_1+ ...... + a_n}n$

In the event that the $a_i$ are all equal, they are all equal to $A$ and $\sqrt[n]{a_1a_2.....a_n} = \sqrt[n]{A^n} = A = \frac {a_1+ ...... + a_n}n$ and we are done.

If they aren't equal then at least one of them is $>A$ (and I'll assume that one that is is $a_1$) and one of the (I'll assume it is $a_2$) is $< A$.

Now if I replace $a_1$ with $A$ and $a_2$ with $a'_2= A-a_1$ we get that the sum $A + a'_2 + a_3 + .... + a_n=a_1 + a_2 + a_3+.. + a_n$ stays the same but tha product $A*a'_2a_3....a_n$ does not.

And if we let $a_1 = A-h$ and $a_2= A+k$ and $a'_2 = A-h+k$ where $h,k>0$ then we have $a_1a_2 = (A-h)(A+k) = A^2-hA+kA - hk$ and $Aa'_2 = A(A-h+k) = A^2-hA+kA = a_1a_2 - hk < a_1a_2$.

So we get $a_1a_2 < Aa'_2$ and therefore $a_1a_2a_3....a_n < Aa'_2a_3....a_n$.

Now if $a'_2 = a_i = A$ we are again done but if not one of $a'_2, a_i$ is $< A$ and another is $> A$. If so let's relabel $a'_2,a_i$ as $b_2,....., b_n$ so that $b_2 < A$ and $b_3 > A$ and we do the same argument we did above and get that

$a_1a_2a_3a_4.......a_n< Aa'_2a_3a_4....a_n= Ab_2b_3b_4.....b_n< A*A*b'_3b_4....b_n$.

We keep repeating the argument until we get

$a_1a_2a_3a_4.......a_n< Aa'_2a_3a_4....a_n< A*A*b'_3b_4....b_n < A*A*A*c'_3c_4...c_4 < ...... < A*A*A*.... *A = A^n$

So we have

$a_1a_2a_3a_4..... a_n\le A^n$ (with equality holding if and only if all $a_i = A$.

$\sqrt[n]{a_1a_2a_3a_4..... a_n}\le \sqrt[n]{A^n} = A=\frac{a_1+a_2 + .....+a_n}n$.

And that's the proof.