# Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2$ where a, b, c and d are positive real numbers

I have to prove the following inequality using the Cauchy-Schwarz inequality: $$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2$$ where a, b, c and d are positive real numbers.

• This is the case $n=4$ of the notorious Shapiro inequality:$$\sum_{i=1}^n\frac{x_i}{x_{i+1}+x_{i+2}}\geqslant\frac{n}{2}\quad (\text{indices modulo } n),$$ which is true only for even $n\leqslant12$ and odd $n\leqslant23$. If someone had asked me to prove this using the Cauchy-Schwarz inequality, giving no other hints, I would be quite cross with them! The case $n=3$, on the other hand, is much friendlier: see Nesbitt's inequality. Mar 6, 2019 at 16:30

By C-S and AM-GM we obtain: $$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c+d)^2}{\sum\limits_{cyc}(ab+ac)}=2+\frac{(a+b+c+d)^2-2\sum\limits_{cyc}(ab+ac)}{\sum\limits_{cyc}(ab+ac)}=$$ $$=2+\frac{a^2+c^2+b^2+d^2-2ac-2bd}{\sum\limits_{cyc}(ab+ac)}\geq2+\frac{2\sqrt{a^2c^2}+2\sqrt{b^2d^2}-2ac-2bd}{\sum\limits_{cyc}(ab+ac)}=2.$$

Use $$xy\leq {(x+y)^2\over 4}$$

if $$x,y\geq 0$$, so

$$\frac{a}{b+c}+\frac{c}{d+a} = {a(a+d)+c(b+c)\over (a+d)(b+c)} \geq 4{a^2+c^2+ad+bc\over (a+b+c+d)^2}$$

and similary $$\frac{b}{c+d}+\frac{d}{a+b}\ge 4{b^2+d^2+ab+dc\over (a+b+c+d)^2}$$

So $$...\geq 4{a^2+c^2+ad+bc+b^2+d^2+ab+dc\over (a+b+c+d)^2}\geq 2$$

A simple proof using convexity of the function $$f(x)=1/x$$. Let S be the sum $$S=a+b+c+d$$ and set $$\tilde{a}=a/S$$ and likewise for b,c,d. Then the inequality is equivalent to : (we drop the tilde's)$$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\geq2$$ where $$a+b+c+d=1$$. Use now the convexity inequality and get that the l.h.s. of the inequality is greater or equal than $$1/({a(b+c)+b(c+d)+c(d+a)+d(a+b)})$$. Then writing the nominator as $$1=(a+b+c+d)^{2}$$ we get that the inequality is obviously true since it is equivalent to $$(a-c)^{2}+(b-d)^{2}\geq 0$$ !!!