let $a,b,c,d,e\ge 0$ be different numbers, show that $$\dfrac{a}{|b-c|}+\dfrac{b}{|c-d|}+\dfrac{c}{|d-e|}+\dfrac{d}{|e-a|}+\dfrac{e}{|a-b|}\ge 3.$$

It seem like Shapiro's inequality $n=5$ case ? Prove of Nesbitt's inequality in 6 variables But I can't use this reslut to prove it.


1 Answer 1


Let $e=\min\{a,b,c,d,e\}$.

Thus, $$\sum_{cyc}\frac{a}{|b-c|}=\frac{a}{|b-c|}+\frac{b}{|c-d|}+\frac{c}{d-e}+\frac{d}{a-e}+\frac{e}{|a-b|}\geq$$ $$\geq \frac{a}{|b-c|}+\frac{b}{|c-d|}+\frac{c}{d}+\frac{d}{a}.$$ Now, let $c=\min\{b,c,d\}$.

Thus, by AM-GM $$\frac{a}{|b-c|}+\frac{b}{|c-d|}+\frac{c}{d}+\frac{d}{a}=\frac{a}{b-c}+\frac{b}{d-c}+\frac{c}{d}+\frac{d}{a}\geq\frac{a}{b}+\frac{b}{d}+\frac{d}{a}\geq3.$$

Let $b=\min\{b,c,d\}$.

Thus, by AM-GM again: $$\frac{a}{|b-c|}+\frac{b}{|c-d|}+\frac{c}{d}+\frac{d}{a}=\frac{a}{c-b}+\frac{b}{|c-d|}+\frac{c}{d}+\frac{d}{a}\geq\frac{a}{c}+\frac{c}{d}+\frac{d}{a}\geq3.$$ Let $d=\min\{b,c,d\}$.

This case for you.

  • $\begingroup$ Nice !+1 Thanks,and I guess for $2n-1$ it's $n$,But I can't prove it $\endgroup$
    – math110
    Aug 24, 2019 at 16:06
  • $\begingroup$ @inequality You are welcome! $\endgroup$ Aug 24, 2019 at 16:07

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