Let $a_{i}\in R(i=1,2,\cdots,n),n\ge 3$. such that $a_{i}\ge \dfrac{n}{2-n},\forall i=1,2,\cdots,n$,and $$a_{1}+a_{2}+\cdots+a_{n}=n$$ show that $$\sum_{i=1}^{n}\dfrac{1}{a^2_{i}}\ge\sum_{i=1}^{n}\dfrac{1}{a_{i}}$$
my attempt
if $n=3$,then $a_{1}+a_{2}+a_{3}=3,a_{i}>-3$,so we must prove $$\sum (a_{1}a_{2})^2\ge \sum (a_{1}a_{2})$$ since $$\sum(a_{1}a_{2})^2\ge a_{1}a_{2}a_{3}(a_{1}+a_{2}+a_{3})=3a_{1}a_{2}a_{3}$$
Let $f(x)=-\frac{1}{x^2}+\frac{1}{x}.$
Thus, we need to prove that $$\sum_{k=1}^nf(a_k)\leq nf\left(\frac{\sum\limits_{k=1}^na_k}{n}\right).$$ But $$f''(x)=\frac{2(x-3)}{x^4}<0$$ for all $x<\frac{\sum\limits_{k=1}^na_k}{n}=1,$
which says that by Vasc's LCF Theorem it's enough to prove our inequality for $a_1=a_2=...=a_{n-1}=a$.
Id est, it's enough to prove that $$(n-1)\left(\frac{1}{a^2}-\frac{1}{a}\right)+\frac{1}{(n-(n-1)a)^2}-\frac{1}{n-(n-1)a}\geq0$$ or $$(a-1)^2(n-(n-2)a)\geq0,$$ which is true because $$n-(n-1)a\geq\frac{n}{2-n}$$ it's $$(n-1)a\leq\frac{n(n-1)}{n-2}$$ or $$n-(n-2)a\geq0$$ and we are done!
Actually, LCF Theorem it's Left Concave Function Theorem by Vasile Cirtoaje.