Có $\frac{a^3}{(b+2c)^2}+\frac{b+2c}{27}+\frac{b+2c}{27}\geq \frac{a}{3}\Rightarrow \frac{a^3}{(b+2c)^2}\geq \frac{9a-2b-4c}{27}$Cmtt: $\frac{b^3}{(c+2a)^2}\geq \frac{9b-2c-4a}{27};\frac{c^3}{(a+2b)^2}\geq \frac{9c-2a-4c}{27}$
Cộng 2 vế 3 bđt trên ta được đpcm