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?
 A: 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.
A: 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?
