Prove Inequality IMO-2012: a2a3⋯an=1

[mathworker]

IMO-2012:
let \(\displaystyle a_2,a_3,...,a_n\) be positive real numbers that satisfy a_2.a₃...a_​n=1 .Prove that,
\(\displaystyle (a_2+1)^2.(a_3+1)^3...(a_n+1)^n>n^n\)
hint:

Spoiler

Use A.M>G.M

 

[mathworker]

Re: prove inequality

Look at this only after giving a serious try
hint#2

Spoiler

split the terms in (a_k+1) into k terms and apply AM>GM

 

[mathworker]

OFICIAL answer by IMO:

Spoiler

\(\displaystyle (a_k+1)=(a_k+\frac{1}{k-1}+\frac{1}{k-1}...\frac{1}{k-1})\)(k-1) times
Apply AM>GM
\(\displaystyle (a_k+1)^k>k^k\frac{a_k}{(k-1)^{k-1}}\)
therefore,
\(\displaystyle \prod_{k=2}^{n} (a_k+1)^k>\prod_{k=2}^{n}k^k\frac{a_k}{(k-1)^{k-1}}\)
\(\displaystyle \prod_{k=2}^{n} (a_k+1)^k>2^2*\frac{a_2}{1^{1}}*3^3*\frac{a_3}{2^{2}}...n^n*\frac{a_{n}}{(n-1)^{n-1}}\)
\(\displaystyle \prod_{k=2}^{n} (a_k+1)^k>\frac{\cancel{2^2}}{1^{1}}\frac{\cancel{3^3}}{\cancel{2^{2}}}...\frac{n^n}{\cancel{(n-1)^{n-1}}}({a_2}.{a_3}...{a_{n}})\)
\(\displaystyle \prod_{k=2}^{n} (a_k+1)^k>n^n.(1)\)