# Proof of an interesting inequality

I think this question was asked here before, but I am unable to find it at the moment. Apologies if this is due to my ineptitude.

Anyway, the question is as follows: let $$n>1$$ be an integer number and $$a_1,\dots,a_n\in\mathbb{R}^+$$. We define $$S_1:=\sum_{i=1}^n a_i$$ and $$S_2:=\sum_{i=1}^n a_i^2$$. Is it true that $$\sum_{i=1}^n\frac{S_1-a_i}{S_2-a_i^2}\geq n\frac{S_1}{S_2}?$$

I am pretty sure it is (basically by qualitative considerations and by the fact that $$\sum_{i=1}^n\frac{S_1-a_i}{S_2-a_i^2}> (n-1)\frac{S_1}{S_2}$$ is trivial), but I was thus far unable to find a proof.

• Did you try a proof by induction? – P3rs3rk3r Mar 19 at 12:33
• Yes, but it didn't seem particularly convenient to me (you only have information on one term of the sum basically). – Leo163 Mar 19 at 12:46
• Anyway, it may be useful to prove the inequality for $n=2, n=3$ before trying the general case. – Vasya Mar 19 at 13:32
• Ignore the proof which i provided some seconds ago, if you are reading it, its wrong. I will adjust it. – P3rs3rk3r Mar 19 at 13:54
• @Vasya For $n=3$ and $n=2$ it's obvious. For $n=4$ it's not so trivial already. – Michael Rozenberg Mar 19 at 19:21

Let $$\frac{S_2}{S_1}=x$$.
Thus, we need to prove that: $$\sum_{i=1}^n\left(\frac{S_1-a_i}{S_2-a_i^2}-\frac{S_1}{S_2}\right)\geq0$$ or $$\sum_{i=1}^n\frac{S_1a_i^2-S_2a_i}{S_2-a_i^2}\geq0$$ or $$\sum_{i=1}^n\frac{a_i^2-xa_i}{xS_1-a_i^2}\geq0$$
or since $$\sum_{i=1}^n(a_i^2-xa_i)=0$$ and $$S_1>x,$$ we need to prove that $$\sum_{i=1}^n\left(\frac{a_i^2-xa_i}{xS_1-a_i^2}-\frac{a_i^2-xa_i}{xS_1-x^2}\right)\geq0$$ or $$\sum_{i=1}^n\left(\frac{1}{xS_1-a_i^2}-\frac{1}{xS_1-x^2}\right)(a_i^2-xa_i)\geq0$$ or $$\sum_{cyc}\frac{(a_i^2-x^2)(a_i^2-xa_i)}{xS_1-a_i^2}\geq0$$ or $$\sum_{cyc}\frac{(a_i-x)^2(a_i+x)a_i}{xS_1-a_i^2}\geq0.$$ Done!