Ta có $\sqrt[3]{1.1.(a+2b)}\leq \frac{a+2b+2}{3}$
$VT=\sum \frac{a^{2}}{\sqrt[3]{a+2b}}\geq \sum \frac{3a^{2}}{a+2b+2}=3(\sum \frac{a^{2}}{a^{2}+2ab+2a})\geq 3\frac{(a+b+c)^{2}}{(a+b+c)^{2}+2(a+b+c)}=3.\frac{1}{3}=1.$
Dấu $"="$ xảy ra $\Leftrightarrow a=b=c=\frac{1}{3}$
- nguyencuong123 yêu thích