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