show this inequality $\sum_{i=1}^{n}\frac{a_{i}a_{i+1}}{1-(a_{i+1}-a_{i})^2}\le \dfrac{1}{2}$

Question

let $n\ge 2$ integer, and $a_{i}\ge 0,i=1,2,\cdots,n$ such $a_{1}+a_{2}+\cdots+a_{n}=1$,show that

$$\sum_{i=1}^{n}\dfrac{a_{i}a_{i+1}}{1-(a_{i+1}-a_{i})^2}\le\dfrac{1}{2}$$ where $a_{n+1}=a_{1}$.

I want to put the denominator $1-(a_{i+1}-a_{i})^2\ge A$ then $$\sum_{i=1}^{n}\dfrac{a_{i}a_{i+1}}{1-(a_{i+1}-a_{i})^2}\le\dfrac{1}{A}\sum_{i}^{n}a_{i}a_{i+1}\le A\dfrac{1}{4A}\left(\sum_{i=1}^{n}a_{i}\right)^2=\dfrac{1}{4A}$$ if we find a best $A\ge \dfrac{1}{2}?$

where use well know inequality
$$(x_{1}+x_{2}+\cdot+x_{n})^2\ge 4(x_{1}x_{2}+x_{2}x_{3}+\cdots+x_{n}x_{1})$$

Maximize $x_1x_2+x_2x_3+\cdots+x_nx_1$

Answer 1

$$\sum_{i=1}^n\frac{a_ia_{i+1}}{1-(a_i-a_{i+1})^2}\leq\sum_{i=1}^n\frac{a_i+a_{i+1}}{4}=\frac{1}{2}.$$ We used the following inequality.

let $a$ and $b$ be non-negative numbers such that $1-(a-b)^2\neq0$ and $a+b\leq1$. Prove that: $$\frac{ab}{1-(a-b)^2}\leq\frac{a+b}{4}.$$

Indeed, we need to prove that: $$4ab\leq a+b-(a+b)(a-b)^2$$ or $$4ab+(a+b)(a-b)^2\leq a+b,$$ for which it's enough to prove that $$4ab+(a-b)^2\leq a+b$$ or $$(a+b)^2\leq a+b$$ or $$a+b\leq1$$ and we are done!