# Prove inequality $\frac{a_1a_2...a_n}{(a_1+a_2+...+a_n)^n}\le \frac{(1-a_1)(1-a_2)...(1-a_n)}{(n-a_1-a_2-...-a_n)^n}$

Let $n\in \mathbb N, a_1,a_2, ...,a_n\in \left(0,\frac 12 \right]$. Prove inequality: $$\frac{a_1a_2...a_n}{(a_1+a_2+...+a_n)^n}\le \frac{(1-a_1)(1-a_2)...(1-a_n)}{(n-a_1-a_2-...-a_n)^n}$$

My work so far:

By induction:

If $n=2$ then $$\frac{ab}{(a+b)^2}\le \frac{(1-a)(1-b)}{(2-a-b)^2} \Leftrightarrow$$ $$\Leftrightarrow4ab-4ab(a+b)+ab(a+b)^2\le ab(a+b)^2-(a+b)^3+(a+b)^2 \Leftrightarrow$$ $$\Leftrightarrow(1-a-b)(a-b)^2 \ge0 -$$obviously, becouse $a,b\in \left(0,\frac 12 \right]$.

• Why didn't you continue your induction proof? May 5 '16 at 10:51
• @YuriyS: I can not do the induction step. May 5 '16 at 10:54

The function $$f(x) = \log\left(\frac{x}{1-x}\right)$$ is concave on $(0,\tfrac12]$ (since $f''\leq 0$ on that interval) so by Jensen's inequality $$\frac{f(a_1)+\ldots+f(a_n)}{n}\leq f\left(\frac{a_1+\ldots+a_n}n\right).$$ Your inequality is an immediate consequence of this.