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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.

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

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!

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