\(\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}\ge\frac{\left(a+b+c\right)^2}{a+b+c+2\left(a^2+b^2+c^2\right)}=\frac{9}{3+2\left(a^2+b^2+c^2\right)}\)
\(\ge\frac{9}{3+2\cdot\frac{\left(a+b+c\right)^2}{3}}=\frac{9}{3+2\cdot\frac{3^2}{3}}=\frac{9}{3+6}=1\)
Dấu bằng xảy ra khi : \(\int^{\frac{a}{a+2b^2}=\frac{b}{b+2c^2}=\frac{c}{c+2a^2}}_{a=b=c}\Rightarrow a=b=c=1\)