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

Question

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.

But I am not able to do it, I am hitting dead-ends with every method I try. Please help!

Answer 1

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

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

Answer 3

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 $(a-c)^{2}+(b-d)^{2}\geq 0$ !!!