khonggilakhongthe

$\sum \left [ \left ( a+1 \right )^2+b^2+1 \right ]\geq \sum ( 2ab+2a+1 )$

===> $2S\leq \sum \frac{1}{ab+a+1}=\frac{1}{ab+a+1}+\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}=1$   (vì $abc=1$)

$"="\Leftrightarrow a=b=c=1$