Ta dễ có đẳng thức: $\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{b+c}{b-c}.\frac{c+a}{c-a}+\frac{c+a}{c-a}.\frac{a+b}{a-b}=-1$
Ta luôn có: $(\frac{a+b}{a-b}+\frac{b+c}{b-c}+\frac{c+a}{c-a})^2\geq 0$
$\Leftrightarrow \sum (\frac{a+b}{a-b})^2+2(\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{b+c}{b-c}.\frac{c+a}{c-a}+\frac{c+a}{c-a}.\frac{a+b}{a-b})\geq 0$
Vậy $\sum (\frac{a+b}{a-b})^2\geq 2(Q.E.D)$
Đẳng thức xảy ra khi $\frac{a+b}{a-b}+\frac{b+c}{b-c}+\frac{c+a}{c-a} = 0$
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