ithanhlam

$\sum \frac{1}{ab+a+1}=\frac{abc}{ab+a+abc}+\frac{1}{bc+b+1}+\frac{b}{cab+cb+b}=\frac{bc}{b+1+bc}+\frac{1}{bc+b+1}+\frac{b}{bc+b+1}=1$

$\Rightarrow \sum \frac{1}{(a+1)^2+b^2+1}=\sum \frac{1}{a^2+b^2+2a+2} \leq \frac{1}{2} \sum \frac{1}{ab+a+1}=\frac{1}{2}$( BĐT Côsy)

Dùng Cauchy á ? 

Cho a, b, c > 0 thỏa mãn abc = 1. Chứng minh rằng: