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Let $x_{i}>0(i=1,2,\cdots,n),n\ge 3$. prove $$\sum_{i=1}^{n}\dfrac{1}{S-x_{i}}+\dfrac{n^n\displaystyle\prod_{i=1}^{n}x_{i}}{(n-1)(n-2)S^n}\ge\dfrac{n-1}{n-2},~~~~~~~~~S=\sum_{i=1}^{n}x_{i}$$

I try to prove Find a function like this$f$ such $$\sum_{i=1}^{n}\dfrac{1}{S-x_{i}}\ge f\left(\dfrac{S^n}{\displaystyle\prod_{i=1}^{n}x_{i}}\right)$$,then I want use $AM-GM$ inequality to prove it. But Until now, I couldn't find this function.

math110
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    As stated like this, the inequality does not hold in general. If all $x_i$ are equal to, say, $a > 0$, then the inequality reduces to $a \leq 1$. Perhaps there is a missing $x_i$ in the numerator of the first sum? – P. Senden Apr 12 '18 at 16:04

2 Answers2

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Expanding on the answer in the comments, a counterexample with $x_i > 0$ is given by $x_i = a$. Then $S = an$, and we have

$$\sum_{i=1}^{n}\dfrac{1}{an-a}+\dfrac{n^na^n}{(n-1)(n-2)a^nn^n}\ge\dfrac{n-1}{n-2}$$ $$\iff\dfrac{n}{a(n-1)}+\dfrac{1}{(n-1)(n-2)}\ge\dfrac{n-1}{n-2}$$ $$\iff\dfrac{n(n-2)}{a}+1\ge(n-1)^2$$ $$\iff n(n-2)+a\ge a(n-1)^2\\ \iff a \leq 1$$ since $n > 2$. So, picking any $a > 1$, say, $a = 2$ gives a counterexample.

B. Mehta
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P. Senden have written in the comments, that the issue inequality is wrong.

The counterexample is $$x= \left\{1, 2, 2\right\},\quad \dfrac{11}{12} < 2.$$

Yuri Negometyanov
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