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$\geq \frac{4(a+c)}{a+b+c+d} + \frac{4(b+d)}{a+b+c+d} = 4$
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#298226 CMR $\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+...
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trong 05-02-2012 - 20:38
$\Leftrightarrow (a+c)(\frac{1}{b+a}+\frac{1}{d+c}) + (b+d)(\frac{1}{b+c}+\frac{1}{d+a})$
$\geq \frac{4(a+c)}{a+b+c+d} + \frac{4(b+d)}{a+b+c+d} = 4$
$\geq \frac{4(a+c)}{a+b+c+d} + \frac{4(b+d)}{a+b+c+d} = 4$
