How prove this $\sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}\frac{a_{i}}{1+a_{i}}\ge n\cdot\prod_{i=1}^{n}a_{i}\sum_{i=1}^{n}\frac{1}{1+a_{i}}$ let $n>2$,and $a_{i}\in(0,1],i=1,2,3,\cdots$,show that
$$\left(\sum_{i=1}^{n}a_{i}\right)\cdot\left(\sum_{i=1}^{n}\dfrac{a_{i}}{1+a_{i}}\right)\ge
n\cdot\left(\prod_{i=1}^{n}a_{i}\right)\cdot\left(\sum_{i=1}^{n}\dfrac{1}{1+a_{i}}\right)$$
This problem is simlar this:Prove $\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}}\ge\sum_{i=1}^{n}\frac{1-a_{i+1}}{1-a_{i}}$ if $a_{i}>0$ and $a_{1}+a_{2}+\cdots+a_{n}=1$
and I guess my this problem can use AM-GM inequality?
Thank you very much
 A: Expand the product of the sums on the left, yielding $\sum_{i,j} \frac{a_ia_j}{1+a_i}$. Except for the summands of the form $\frac{a_i^2}{1+a_i}$, you can use the fact that $a_ia_j \geq \prod_k a_k$ to bound each summand $\frac{a_ia_j}{1+a_i}$ on the left by a corresponding summand $ \frac{\prod_k a_k}{1+a_i}$ on the right.
After this reduction, it will suffice to prove that: $$\sum_{i} \frac{a_i^2}{1+a_i} \geq \sum_i \frac{\prod_j a_j}{a_i+1}.$$
For $n = 2 $ this says simply: $\frac{a^2}{1+a} + \frac{b^2}{1+b} \geq ab( \frac{1}{1+a} + \frac{1}{1+b})$. You can prove this by brute force. Multiply both sides by $(1+a)(1+b)$, so that it will suffice to show that:
$$a^2 +b^2 + ab(a+b) = a^2(1+b) + b^2(1+a) \geq ab(1+b+1+a) = ab(a+b) + 2ab,$$
which is clear because $a^2 + b^2 \geq 2ab$.
For $n>2$, apply the case $n=2$ to each (unordered) pair $i,j$ with $i \neq j$, take the sum of resulting inequalities and divide by $n-1$ (that's the number of pairs containing any given $i$). This leads to:
 $$\sum_{i} \frac{a_i^2}{1+a_i} \geq \frac{1}{n-1} \sum_{i \neq j} \frac{a_i a_j}{a_i+1} \geq \frac{1}{n-1}  \sum_{i \neq j} \frac{\prod_k a_k}{a_i+1} = \sum_i \frac{\prod_k a_k}{a_i+1}.$$
This is precisely what was to be proved.
