\(VT=\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\)
\(=\frac{a^2}{ab+2ac}+\frac{b^2}{bc+2ab}+\frac{c^2}{ac+2bc}\)
Áp dụng bđt Cauchy-Schwarz dạng Engel:
\(\frac{a^2}{ab+2ac}+\frac{b^2}{bc+2ab}+\frac{c^2}{ac+2bc}\ge\frac{\left(a+b+c\right)^2}{ab+2ac+bc+2ab+ac+2bc}\)\(=\frac{\left(a+b+c\right)^2}{3\left(ab+ac+bc\right)}\)\(=\frac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=1\)