∑ 1/(1+ab)≥9/(3+ab+bc+ca)≥9/(3+(a+b+c)^2/3)=27/13.

wer, are you sure?

ThirdTimeLucky wrote: Let $ a,b,c $ be positive real numbers satisfying $ a+b+c=2 $. Prove the inequality \[ \frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ca} \ge \frac{27}{13} \] $uvw$ kills it immediately.

Hi!! Please tell if I am wrong! $\sum\frac{1}{1+ab}\geq\frac{9}{3+\sum ab}$ Thus we have to prove: $\frac{9}{3+\sum ab}\geq\frac{27}{13}$ Which is equivalent to $\sum ab\leq\frac{4}{3}$ Which follows directly from $\sum a^2\geq 3\sum ab$ EDIT: Sorry didn't read wer's solution!! It is the same!!