# Proving $(1+a_{2})^{2}(1+a_{3})^{3}\cdots(1+a_{n})^{n}\ge n^n$ for positive reals $a_2,\ldots, a_n$ whose product is $1$

Let $n \ge3$ be an integer, and let $a_{2},a_{3}, ... ,a_{n}$ be positive real numbers such that $a_{2} a_{3}\cdots a_{n}=1.$ Prove that: $$(1+a_{2})^{2}(1+a_{3})^{3}\cdots(1+a_{n})^{n}\ge n^n$$

This is the 2nd problem of the 53rd IMO and seems pretty interesting. How would we solve that?

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@ Angela Richardson: thanks for that link. The proof is very nice (marvellous)! –  Chris's sis Jul 21 '12 at 16:36
In fact, all proofs there are nice and simple. –  Chris's sis Jul 21 '12 at 17:01
Both displayed equations in the proof on the imomath site are marred by a serious, different, misprint. –  Did Jul 21 '12 at 17:07
@ did: that's true, but i've got the main idea proof. I do mistakes, as well. I wonder if there could also be other nice approaches ... –  Chris's sis Jul 21 '12 at 17:12

Set $a_2 =\frac{x_2}{x_3}, a_3=\frac{x_3}{x_4},\ldots, a_n=\frac{x_n}{x_2}$. This is a very useful substitution that we use in cases when we have a product equal to one like in this one $a_2 a_3 \cdots a_n=1$.
Now we need to prove that $$(x_2+x_3)^2 (x_3+x_4)^3 \cdots (x_n+x_2)^n > n^n x_3^2 x_4^3 \cdots x_{n}^{n-1}x_2^n$$
which become obvious since for each $k$ by applying the Arithmetic-Geometric Mean we have that: $$(x_k+x_{k+1})^{k}=\left(x_k+(k-1)\frac{x_{k+1}}{k-1}\right)^k\geqslant k^k x_k\frac{x_{k+1}^{k-1}}{(k-1)^{k-1}}$$
Just multiply for $k$ from $2$ to $n$.