Áp dụng BĐT Bunyakovsky dạng phân thức ta có:
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{bc+ba}+\frac{c^2}{ca+cb}\)
\(\ge\frac{\left(a+b+c\right)^2}{ab+bc+bc+ca+ca+ab}=\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
Mà \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{\left(a+b+c\right)^2}{\frac{2\left(a+b+c\right)^2}{3}}=\frac{3}{2}\)
Dấu "=" xảy ra khi: a = b = c