# Equality in AM GM Inequality

In AM GM inequality for nonnegative real numbers $a_1,a_2,\ldots,a_n$, How to show that if equality holds then $a_1=a_2=\ldots=a_n,$ using method of induction?

• It is obviously true when n=1. Now assume it is true for some $n$, and use that to prove it is also true for $n+1$. – Paul Sinclair Sep 2 '15 at 4:36
• I can't imagine that they want you to use induction unless they're asking for the case in which $n$ is a power of $2$. – Omnomnomnom Sep 2 '15 at 4:51
• This usually comes out of the proof of the inequality. What proof do you have of that? – mickep Sep 2 '15 at 5:20

Edit Sorry, I read now that you wanted a proof using induction. The proof below does not. I let it stand, since I think it is a nice proof, and someone might thing it is helpful.

For two terms, the statement follows by expanding $$0\leq (a_1-a_2)^2,$$ and noting that the inequality is an equality precisely when $a_1=a_2$.

Now suppose, to get a contradiction, that not all $a_i$, $1\leq i\leq n$ are equal and that the AM equals the GM, i.e. $$\frac{a_1+a_2+a_3+\cdots+a_n}{n}=(a_1a_2\cdots a_n)^{1/n}.$$ We can without loss of generality assume that $a_1\neq a_2$. Then, if we replace $a_1$ and $a_2$ by their arithmetic mean, $$\tilde{a}_1=\frac{a_1+a_2}{2},\quad \tilde{a}_2=\frac{a_1+a_2}{2},$$ we get $n$ numbers $\tilde{a}_1$, $\tilde{a}_2$, $a_3$, $\ldots$, $a_n$, that has the same arithmetic mean as the original one, but what happens with the geometric mean? From the AM-GM inequality with two terms $$\tilde{a}_1\tilde{a}_2=\Bigl(\frac{a_1+a_2}{2}\Bigr)^2 > a_1a_2.$$ Note the strict inequality (which is a result of the case of two terms). Hence, $$\frac{\tilde{a}_1+\tilde{a}_2+a_3+\cdots+a_n}{n}= \frac{a_1+a_2+a_3+\cdots+a_n}{n}=(a_1a_2\cdots a_n)^{1/n}<(\tilde{a}_1\tilde{a}_2a_3\cdots a_n)^{1/n}$$ contradicting the AM-GM inequality.

• Brilliant! Thank you! – Deepak Gupta Sep 2 '15 at 7:47
• Just for the sake of completeness, should you also add one step in the beginning that AM is equal to GM if all numbers are equal for any value of n? I do know it's trivial but, as I said, just for the sake of completeness. – Deepak Gupta Sep 2 '15 at 7:51
• @DeepakGupta I agree that one could add that as a comment, but logically that is not what we need to prove. – mickep Sep 2 '15 at 8:13

Requesting everyone to kindly point out the flaw in my reasoning in the comments to this answer..

$n=1$ and $n=2$ are easy to prove cases. Assuming the hypothesis to be true for $n=k$ where $k>2$ we get $AM=GM=a_1$

For $n=k+1$, we can create $(k+1)$ subsets of $k$ numbers, for each of which, the hypothesis implies that all elements of the subsets are equal as long as $AM=GM=b_i$ without any need to worry about whether $b_i=a_1$ or not for any $i$ between $1$ and $k+1$.

Note that any two subsets differ in just one element and hence have $k-1$ common elements. Equality of all elements in two such subsets would imply $a_1=a_2=...=a_{k+1}$ which is the desired result.

This seems to be a reasonable line of thought to me. Am I missing something?

• The only missing link is that your induction step doesn't work for $k=1, n=2$ as the two subsets of size 1 are disjoint. – IanF1 Sep 2 '15 at 5:22
• Thanks, Ian. I improved my answer to handle this discrepancy. However, I think there is another flaw. Given that the $(k+1)$ numbers have $AM=GM$, it would not necessarily imply that $AM=GM$ for every subset of $k$ elements. My solution requires that EVERY such subset MUST have $AM=GM$ if that is true of the superset. – Deepak Gupta Sep 2 '15 at 5:35
• Quite so. And you should start by assuming it holds for some $n \geq 2$ ,not n=k>2 – DanielWainfleet Sep 2 '15 at 7:03
• That is not required, since I have taken two special cases (n=1 and n=2) (instead of using one standard base case) and proved them separately.. Hence, reasoning for all n=k>2 is sufficient to complete the proof – Deepak Gupta Sep 2 '15 at 7:26