for positive reals a,b,c,d

show that

$\displaystyle \frac{b(c+a)}{c(a+b)}+\frac{c(b+d)}{d(b+c)}+\frac{ d(c+a)}{a(c+d)}+\frac{a(d+b)}{b(d+a)}\geq 4$

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- Aug 10th 2010, 09:32 PMsmartyhelp me with this inequality
for positive reals a,b,c,d

show that

$\displaystyle \frac{b(c+a)}{c(a+b)}+\frac{c(b+d)}{d(b+c)}+\frac{ d(c+a)}{a(c+d)}+\frac{a(d+b)}{b(d+a)}\geq 4$