matinyousefi wrote:
For any positive numbers $ a_1 , a_2 , \dots, a_n $ prove the inequality $$\!
\left(\!1\!+\!\frac{1}{a_1(1+a_1)} \!\right)\!
\left(\!1\!+\!\frac{1}{a_2(1+a_2)} \! \right) \! \dots \!
\left(\!1\!+\!\frac{1}{a_n(1+a_n)} \! \right) \geq
\left(\!1\!+\!\frac{1}{p(1+p)} \! \right)^{\! n} \! ,$$where $p=\sqrt[n]{a_1 a_2 \dots a_n}$.
Take natural log in both sides and let $f(x)=1+ \frac{1}{e^x(1+e^x)}$. Moreover, let $b_i = \ln a_i$. Therefore we must prove:
$$ f(b_1)+f(b_2)+...+f(b_n) \ge n f( \frac{b_1+...+b_n}{n})$$
We can compute $f'(x)=-e^{-x}(1+e^x)^{-1}-(1+e^x)^{-2}$ and $f''(x)=e^{-x}(1+e^x)^{-1}+(1+e^x)^{-2}+2e^x(1+e^x)^{-3}$ that is clearly $\ge 0$. Hence $f$ is convex and we are done by Jensen.