# Prove inequality

#### mathworker

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

#### mathworker

##### Well-known member
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

#### mathworker

##### Well-known member
$$\displaystyle (a_k+1)=(a_k+\frac{1}{k-1}+\frac{1}{k-1}........\frac{1}{k-1})$$(k-1) times
$$\displaystyle (a_k+1)^k>k^k\frac{a_k}{(k-1)^{k-1}}$$
$$\displaystyle \prod_{k=2}^{n} (a_k+1)^k>\prod_{k=2}^{n}k^k\frac{a_k}{(k-1)^{k-1}}$$
$$\displaystyle \prod_{k=2}^{n} (a_k+1)^k>2^2*\frac{a_2}{1^{1}}*3^3*\frac{a_3}{2^{2}}......n^n*\frac{a_{n}}{(n-1)^{n-1}}$$
$$\displaystyle \prod_{k=2}^{n} (a_k+1)^k>\frac{\cancel{2^2}}{1^{1}}\frac{\cancel{3^3}}{\cancel{2^{2}}}......\frac{n^n}{\cancel{(n-1)^{n-1}}}({a_2}.{a_3}.......{a_{n}})$$
$$\displaystyle \prod_{k=2}^{n} (a_k+1)^k>n^n.(1)$$