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Suppose that $n\geq3$. I want to prove that

$ v_n = \sum\limits_{p=1}^{n-1} \frac{1}{p(n+1-p)(n-p)} \geq \frac{2n+1}{2n^2} .$

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    $\begingroup$ try induction with respect to $n$ $\endgroup$ – Dr. Sonnhard Graubner Jan 19 '17 at 18:50
  • $\begingroup$ "Good" hint ... $\endgroup$ – tired Jan 19 '17 at 18:59
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By partial fraction decmposition (treating $p$ as $x$) $$\frac{1}{p(n+1-p)(n-p)}=\frac{1}{n(n+1)}\cdot\frac{1}{p}+\frac{1}{n}\cdot\frac{1}{n-p}-\frac{1}{n+1}\cdot\frac{1}{n+1-p}$$ hence by summing both sides for $p=1,2,\ldots,n-1$ we get:

$$\sum_{p=1}^{n-1}\frac{1}{p(n+1-p)(n-p)} = \frac{H_{n-1}}{n(n+1)}+\frac{H_{n-1}}{n}+\frac{1-H_n}{n+1}$$ or, by simplifying,

$$\sum_{p=1}^{n-1}\frac{1}{p(n+1-p)(n-p)} = \frac{2H_{n-1}+n-1}{n(n+1)}\geq\frac{n+2}{n(n+1)}\geq\frac{2n+1}{2n^2}.$$

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  • $\begingroup$ Thank you for your help :-) I should have think to partial decomposition... $\endgroup$ – Babyblog Jan 20 '17 at 9:52

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