MHB Prove Inequality IMO-2012: a2a3⋯an=1

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IMO-2012:
let $$a_2,a_3,...,a_n$$ be positive real numbers that satisfy a2.a3...a​n=1 .Prove that,
$$(a_2+1)^2.(a_3+1)^3...(a_n+1)^n>n^n$$
hint:
Use A.M>G.M
 
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Re: prove inequality

Look at this only after giving a serious try
hint#2
split the terms in (ak+1) into k terms and apply AM>GM
 
OFICIAL answer by IMO:
$$(a_k+1)=(a_k+\frac{1}{k-1}+\frac{1}{k-1}...\frac{1}{k-1})$$(k-1) times
Apply AM>GM
$$(a_k+1)^k>k^k\frac{a_k}{(k-1)^{k-1}}$$
therefore,
$$\prod_{k=2}^{n} (a_k+1)^k>\prod_{k=2}^{n}k^k\frac{a_k}{(k-1)^{k-1}}$$
$$\prod_{k=2}^{n} (a_k+1)^k>2^2*\frac{a_2}{1^{1}}*3^3*\frac{a_3}{2^{2}}...n^n*\frac{a_{n}}{(n-1)^{n-1}}$$
$$\prod_{k=2}^{n} (a_k+1)^k>\frac{\cancel{2^2}}{1^{1}}\frac{\cancel{3^3}}{\cancel{2^{2}}}...\frac{n^n}{\cancel{(n-1)^{n-1}}}({a_2}.{a_3}...{a_{n}})$$
$$\prod_{k=2}^{n} (a_k+1)^k>n^n.(1)$$
 
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