# $AM-GM$ Inequality Proof From a Book

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 $$a_{i} and $$a_{j}>A$$. Because if all $$a_{i}$$s $$>A$$ or all $$a_{i}$$s $$ 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.

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

• It's induction. If $A$ is the average of $a_1$ to $a_n$, You can have a series of $a_1 + a_2 + a_3 + ..... + a_n = A + a_2' + a_3 +....a_n$ but $Aa_2'a_3a_4...a_n> a_1a_2a_3a_4...a_n$ and then you do it again to get $A+a_2' + a_3 +.... + a_n= A+A+ a_3' + a_4 + ...+a_n$ but $A*Aa_3'a_4...a_n> Aa_2'a_3a_4...a_n$ and you keep repeating untill you get $a_1a_2...a_n<.....<A*A*....*A$ so $\sqrt[n]{a_1a_2...a_n}<\sqrt[n]{A^n} = a=\frac{a_1 + .....+ a_n}n$. – fleablood Jan 27 at 6:11
• Ah thanks. I should have seen that. – Arief Anbiya Jan 30 at 8:15

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.

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.

• Technically they're $n$-tuples, not sets. – J.G. Jan 27 at 6:30
• But they work better as sets with constant reordering as far as the logic of the original text goes – fleablood Jan 27 at 18:15
• That's why I thought of them as being sets, although J.G. is correct that what I had written before I added the curly brackets were technically $n$-tuples rather than sets. – lonza leggiera Jan 27 at 22:57