Bạn cm bđt Nesbitt như thế này:
Theo Cauchy-Schwartz ta có:[tex][(a+b)+(b+c)+(c+a)][tex]\Leftrightarrow 2(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})\geq 9 \Leftrightarrow (a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})\geq \frac{9}{2}[/tex] [/tex]
[tex]\Leftrightarrow \frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}\geq \frac{9}{2} \Leftrightarrow 1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}\geq \frac{9}{2}[/tex]
[tex]\Leftrightarrow \frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}\geq \frac{9}{2}-1-1-1=\frac{3}{2}[/tex] (đpcm)