cho $a, b, c>0, a+b+c=3$. chứng minh
$\sum_{cyc}{\frac{(a+b)(b+c)}{b^3+c^3+abc}}+\frac{1}{4}\ge\frac{a^2+b^2+c^2+48abc}{4(ab+bc+ca)}$
$\sum_{cyc}{\frac{(a+b)(b+c)}{b^3+c^3+abc}}+\frac{1}{4}\ge\frac{a^2+b^2+c^2+48abc}{4(ab+bc+ca)}$
Started By Gachdptrai12, 22-05-2016 - 11:49
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Posted 22-05-2016 - 11:49
Gachdptrai12
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