\(S=\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}=\frac{a^2}{ab+2ac}+\frac{b^2}{bc+2ab}+\frac{c^2}{ac+2bc}\)
\(\Rightarrow S\ge\frac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge\frac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=1\)
Dấu "=" xảy ra khi \(a=b=c\)