show this inequality if $\sum_{1\le i<j\le n}a_{i}a_{j}=\binom{n}{2}$

Question

let $a_{i}>0,i=1,2,3,\cdots,n$,and such$$\sum_{1\le i<j\le n}a_{i}a_{j}=\binom{n}{2}$$ show that $$\sum_{i=1}^{n}\dfrac{1}{a_{i}+n-1}\le 1$$

it is clear when $n=2$ this $$\dfrac{1}{1+a}+\dfrac{1}{1+b}=1$$ when $ab=1$

Answer 1

It's wrong!

Try $n=4$, $a_4\rightarrow0^+$ and $a_1=a_2=a_3\rightarrow\sqrt2.$

We obtain: $$LSH-1\rightarrow\frac{3}{3+\sqrt2}+\frac{1}{3}-1=0.0129...>0.$$