$\dpi{150} a\geq b\geq c\rightarrow \left\{\begin{matrix} a^{2}\geq b^{2}\geq c^{2} & & \\ \frac{1}{b+2c}\geq \frac{1}{c+2a}\geq \frac{1}{a+2b}& & \end{matrix}\right.\\chebyshev\sum \frac{a^{2}}{b+2c}\geq \frac{1}{3}(\sum a^{2})(\sum \frac{1}{b+2c})\geq \frac{\sum a^{2}}{\sum a}\geq \sqrt{\frac{\sum a^{2}}{3}}=1$
chưa chắc gì $c+2a \le a+2b$