ta có [tex]\frac{a}{1+b-a}+\frac{b}{1+c-b}+\frac{c}{1+a-c}=\frac{a}{c+2b}+\frac{b}{a+2c}+\frac{c}{2a+b}[/tex]
[tex]=\frac{a^2}{2ab+ac}+\frac{b^2}{ab+2bc}+\frac{c^2}{2ac+bc}[/tex]
Áp dụng BĐT Sơ vác ta có [tex]\frac{a^2}{2ab+ac}+\frac{b^2}{2bc+ab}+\frac{c^2}{2ac+bc}\geq \frac{(a+b+c)^2}{3(ab+bc+ac)}[/tex]
Lại có [tex](a+b+c)^2\geq 3(ab+bc+ac)[/tex]
=>ĐPCM
Dấu "="[tex]<=>a=b=c=\frac{1}{3}[/tex]