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o0ranangel0o
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Cho a,b,c >0 thoả mãn ${a}^{2}+{b}^{2}+{c}^{2}=12$. Chứng minh
$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}$\geq $\frac{8}{{a}^{2}+28}+\frac{8}{{b}^{2}+28}+\frac{8}{{c}^{2}+28}$
$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}$\geq $\frac{8}{{a}^{2}+28}+\frac{8}{{b}^{2}+28}+\frac{8}{{c}^{2}+28}$