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KMTTQ, giả sử $a=\min \{ a,b,c\} $. Ta có: $T=\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}-\frac{1}{a(b+c)}-\frac{1}{b(c+a)}-\frac{1}{c(a+b)}$ $=\frac{1}{a+b}\left( \frac{1}{b}-\frac{1}{c}\right) +\frac{1}{b+c}\left( \frac{1}{c}-\frac{1}{a}\right) +\frac{1}{c+a}\left( \frac{1}{a}-\frac{1}{b}\right)$ Nếu $b\leq c$ thì $T\geq \frac{1}{c+a}\left( \frac{1}{b}-\frac{1}{c}\right)+\frac{1}{c+a}\left( \frac{1}{c}-\frac{1}{a}\right)+\frac{1}{c+a}\left( \frac{1}{a}-\frac{1}{b}\right)=0$. Nếu $b>c$ thì $T>\frac{1}{a+b}\left( \frac{1}{b}-\frac{1}{c}\right) +\frac{1}{a+b}\left(
\frac{1}{c}-\frac{1}{a}\right) +\frac{1}{a+b}\left(
\frac{1}{a}-\frac{1}{b}\right)=0$. Tóm lại: $\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\geq \frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)}\geq \frac{9}{2(ab+bc+ca)}$.
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