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