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I want to prove the inequality in a general case ($n\ge3$): $\;\sum\limits_{i=1}^n \dfrac{p_i}{p_{i+1}+p_{i+2}} \ge \dfrac{n}{2}$ for any positive numbers $p_i,\,i=\overline{1,n}$, where $p_{n+1} = p_1$ and $p_{n+2} = p_2$. It seems that there is no counter-example...

In particular cases when $n=3$, $n=4$ and $n=5$ there is no problem with use of Jensen's inequality for the function $f(x) = \dfrac{1}{x}$:

$$\sum\limits_{i=1}^n \dfrac{m_i}{x_i} \ge \dfrac{\left(\sum_{i=1}^n m_i\right)^2}{\sum_{i=1}^n m_i x_i}$$

One can choose $m_i=p_i$ and $x_i=p_{i+1}+p_{i+2}$. Then, for example, for $n=3$ we have:

$$\dfrac{p_1}{p_2+p_3} + \dfrac{p_2}{p_3+p_1} + \dfrac{p_3}{p_1+p_2} \ge \dfrac{(p_1 + p_2 + p_3)^2}{p_1(p_2+p_3) + p_2(p_3+p_1) + p_3(p_1+p_2)}.$$

RHS is equal to

$$\dfrac{(p_1 + p_2 + p_3)^2}{p_1^2 + p_2^2 + p_3^2 + 2(p_1p_2 + p_2p_3 + p_3p_1) - (p_1^2 + p_2^2 + p_3^2)} = \dfrac{(p_1 + p_2 + p_3)^2}{(p_1 + p_2 + p_3)^2 - (p_1^2 + p_2^2 + p_3^2)}.$$

According to Cauchy–Bunyakovsky inequality for $a_1=a_2=a_3=1;\; b_i=p_i$: $$(1^2+1^2+1^2)\cdot (p_1^2 + p_2^2 + p_3^2) \ge (p_1 + p_2 + p_3)^2,$$ finally we have the following:

$$\dfrac{p_1}{p_2+p_3} + \dfrac{p_2}{p_3+p_1} + \dfrac{p_3}{p_1+p_2} \ge \dfrac{(p_1 + p_2 + p_3)^2}{(p_1 + p_2 + p_3)^2 - \frac{1}{3} (p_1 + p_2 + p_3)^2} = \dfrac{3}{2}.$$

But there is a problem with a plenty of $p_i p_j$ in cases starting with $n=6$. Can someone help to prove the inequality for $n=6$ or greater?

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    $\begingroup$ That is Shapiro's inequality. It holds for $n \le 13$ but not in general. $\endgroup$
    – Martin R
    Nov 10, 2019 at 16:19
  • $\begingroup$ @Martin R From Wikipedia, the proof for $n=23$ rely on numerical computations, and an analytical proof may be expected? en.wikipedia.org/wiki/Shapiro_inequality $\endgroup$
    – River Li
    Nov 10, 2019 at 23:40
  • $\begingroup$ @MartinR, thanks! I couldn't find it by myself. $\endgroup$ Nov 12, 2019 at 7:17

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For $n=6$ we need to prove that $$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d}{e+f}+\frac{e}{f+a}+\frac{f}{a+b}\geq3,$$ where $a$, $b$, $c$, $d$, $e$ and $f$ are positives.

Indeed, by C-S we obtain: $$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c+d+e+f)^2}{\sum\limits_{cyc}(ab+ac)}=$$ $$=\frac{(a+d+b+e+c+f)^2}{(a+d)(b+e)+(a+d)(c+f)+(b+e)(c+f)}\geq3.$$

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