# show this inequaliy $\sum_{cyc}\frac{a}{|b-c|}\ge 3$

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

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.

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