Prove the inequality $\left(1+\frac{1}{a_1(1+a_1)}\right)...\left(1+\frac{1}{a_k(1+a_k)}\right)\ge\left(1+\frac{1}{p(1+p)}\right)^k$

Question

Let $a_1, a_2,...,a_k$ are any positive real numbers. Prove the inequality $$\left(1+\frac{1}{a_1(1+a_1)}\right)\left(1+\frac{1}{a_2(1+a_2)}\right)...\left(1+\frac{1}{a_k(1+a_k)}\right)\ge$$ $$\ge\left(1+\frac{1}{p(1+p)}\right)^k$$ where $\sqrt[k]{a_1a_2...a_k}=p$.

My work so far:

I used Jensen's inequality.

Let $f(x)=\ln\left(1+\frac{1}{x(1+x)}\right)$. If $x\ge0$ then $f''(x)\ge0$. Then $$\ln\left(1+\frac{1}{a_1(1+a_1)}\right)+...+\ln\left(1+\frac{1}{a_k(1+a_k)}\right)\ge k\ln\left(1+\frac{1}{q(1+q)}\right)$$ where $q=\frac{a_1+a_2+...a_k}{k}$.

Then $$\ln\left(\left(1+\frac{1}{a_1(1+a_1)}\right)...\left(1+\frac{1}{a_k(1+a_k)}\right)\right)\ge \ln\left(1+\frac{1}{q(1+q)}\right)^k$$ $$\left(1+\frac{1}{a_1(1+a_1)}\right)...\left(1+\frac{1}{a_k(1+a_k)}\right)\ge \left(1+\frac{1}{q(1+q)}\right)^k$$ But $$\left(1+\frac{1}{q(1+q)}\right)\not \ge\left(1+\frac{1}{p(1+p)}\right)$$

I can not finish the proof of this inequality

Addition 1:

Addition 2:

Answer 1

Let $x_i=\ln\frac{a_i}{p}.$

Thus, $\sum\limits_{i=1}^nx_i=0$ and we need to prove that $$\sum_{i=1}^kf(x_i)\geq k\ln\left(1+\frac{1}{p(1+p)}\right),$$ where $$f(x)=\ln\left(1+\frac{1}{pe^x(1+pe^x)}\right).$$ But, $$f''(x)=\frac{p^2e^{2x}(4p^2e^{2x}+7pe^x+4)}{(pe^x+1)^2(p^2e^{2x}+pe^x+1)^2}>0.$$ Id est, by Jensen $$\sum_{i=1}^kf(x_i)\geq kf\left(\frac{\sum\limits_{i=1}^kx_i}{k}\right)=kf(0)=k\ln\left(1+\frac{1}{p(1+p)}\right).$$ Done!

Answer 2

Hint: Set $b_i = \ln(a_i)$ and apply Jensen's inequality to $$\ln\left(1+\frac{1}{e^x(1+e^x)}\right).$$