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Cho $ a,b,c,d \geq 0 $ chứng minh rằng
$$ A = \frac{b(a+c)}{c(a+b)} + \frac{c(b+d)}{d(b+c)} + \frac{d(a+c)}{a(c+d)} + \frac{a(b+d)}{b(d+a)} \geq 4 $$
$$ A = \frac{b(a+c)}{c(a+b)} + \frac{c(b+d)}{d(b+c)} + \frac{d(a+c)}{a(c+d)} + \frac{a(b+d)}{b(d+a)} \geq 4 $$
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