If $a, b, c, d\in\mathbb R^+, $ then prove that $\displaystyle\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}\ge 0.$

Question

My approach: We have:

$\displaystyle\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}$

$\displaystyle=\frac{a+c}{b+c}-1+\frac{b+d}{c+d}-1+\frac{c+a}{d+a}-1+\frac{d+b}{a+b}-1$

$\displaystyle=(c+a)\left(\frac1{b+c}+\frac1{d+a}\right)+(b+d)\left(\frac1{c+d}+\frac1{a+b}\right)-4\ge (c+a) \left(\frac{2\cdot 2}{b+c+d+a}\right)+(b+d)\left(\frac{2\cdot 2}{c+d+a+b}\right)-4$

(on applying AM $\ge$ HM.)

$\displaystyle=\left(\frac4{a+b+c+d}\right) (c+a+b+d)-4=0.$

Hence the inequality.

But I would love to know is there any other elegant method to prove the same? Please mention. Thanks in advance.

Answer 1

We can use C-S one time only: $$\sum_{cyc}\frac{a-b}{b+c}=-4+\sum_{cyc}\left(\frac{a-b}{b+c}+1\right)=-4+\sum_{cyc}\frac{a+c}{b+c}=$$ $$-4+\sum_{cyc}\frac{(a+c)^2}{(b+c)(a+c)}\geq-4+\frac{\left(\sum\limits_{cyc}(a+c)\right)^2}{\sum\limits_{cyc}(b+c)(a+c)}=-4+\frac{4(a+b+c+d)^2}{\sum\limits_{cyc}(ab+ab+ac+a^2)}=0.$$

Answer 2

If $a-b\geq b-c \geq c-d \geq d-a$ and $\frac{1}{b+c}\geq \frac{1}{c+d}\geq \frac{1}{d+a}\geq \frac{1}{a+b}$

We can apply Chebyshev's sum inequality to get :

$$\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}\geq 0.25(a-b+b-c+c-d+d-a)\Big(\frac{1}{b+c}+ \frac{1}{c+d}+\frac{1}{d+a}+\frac{1}{a+b}\Big)=0$$

For the other cases you can apply rearrangement inequality