Cho các số dương a,b,c thoả mãn $(a+b)(b+c)(c+a)=8$.Chứng minh rằng
$\frac{a+b+c}{3}\geq \sqrt[27]{\frac{a^{3}+b^{3}+c^{3}}{3}}$
$\frac{a+b+c}{3}\geq \sqrt[27]{\frac{a^{3}+b^{3}+c^{3}}{3}}$
Started By neversaynever99, 17-10-2013 - 00:57
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