\(\frac{a}{1+b-a}+\frac{b}{1+c-b}+\frac{c}{1+a-c}\)
\(=\frac{a}{2b+c}+\frac{b}{2c+a}+\frac{c}{2a+b}\)
\(=\frac{a^2}{2ab+ac}+\frac{b^2}{2bc+ab}+\frac{c^2}{2ac+bc}\)
Ta có: \(VT\ge\frac{\left(a+b+c\right)^2}{3\left(ab+bc+ac\right)}\ge\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
Ta được đpcm.