I have solve following problem:
let $a_{i}>0,(i=1,2,\cdots,n)$,and such $\sum_{i=1}^{n}a_{i}=1$,show that $$F=\sum_{1\le i<j\le n}\dfrac{a_{i}a_{j}}{(1-a_{i})(1-a_{j})}\ge\dfrac{n}{2n-2}$$ Proof:use Cauchy-Schwarz inequality we have (let $t_{i}=a_{i}(1-a_{i})$) $$4F\ge\dfrac{\left(2\sum_{1\le i<j\le n}a_{i}a_{j}\right)^2}{\displaystyle\sum_{1\le i<j\le n}a_{i}(1-a_{i})a_{j}(1-a_{j})}=\dfrac{(\displaystyle\sum_{i=1}^{n}t_{i})^2}{\displaystyle\sum_{1\le i<j\le n}t_{i}t_{j}}$$ it is enough to prove that $$(n-1)(\sum_{i=1}^{n}t_{i})^2\ge 2n\sum_{1\le i<j\le n}t_{i}t_{j}$$ or $$n\sum_{i=1}^{n}t^2_{i}\ge (\sum_{i=1}^{n}t_{i})^2$$ it is Cauchy-Schwarz inequality!
My problem:I found this problem is also for $a_{i}\in (-1,1)(i=1,2,\cdots,n)$,and $a_{1}+a_{2}+\cdots+a_{n}=1$.we have $$F=\sum_{1\le i<j\le n}\dfrac{a_{i}a_{j}}{(1-a_{i})(1-a_{j})}\ge\dfrac{n}{2n-2}\tag{1}$$
this case we can't use Cauchy-Schwarz inequality to solve it.because this $a_{i}(1-a_{i})a_{j}(1-a_{j})$ maybe is $<0$.so How to prove $(1)$
