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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?

Math_guy
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2 Answers2

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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.

mickep
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  • Brilliant! Thank you! – Deepak Gupta Sep 02 '15 at 07:47
  • 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 02 '15 at 07:51
  • @DeepakGupta I agree that one could add that as a comment, but logically that is not what we need to prove. – mickep Sep 02 '15 at 08:13
  • Ah, exquisite argument. Thank you. I think this argument is a nice complement to Cauchy's AM-GM proof (at least the one in Wikipedia). I was reading the Wikipedia article on the proof, and – perhaps it was the non-math wizard in me – but I couldn't understand why x_i had to be equal for all i in order for the equality to hold. Like, I know that, but couldn't deduce it from the proof. This makes it much clearer! – BitParser Dec 15 '21 at 00:13
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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?

Deepak Gupta
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  • 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 02 '15 at 05:22
  • 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 02 '15 at 05:35
  • Quite so. And you should start by assuming it holds for some $n \geq 2$ ,not n=k>2 – DanielWainfleet Sep 02 '15 at 07:03
  • 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 02 '15 at 07:26