<s>[INST] 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.  

But I am not able to do it, I am hitting dead-ends with every method I try. Please help!
 [/INST] 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 $$...\geq 4{a^2+c^2+ad+bc+b^2+d^2+ab+dc\over (a+b+c+d)^2}\geq 2$$
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