Let $q=ab+bc+ca, p=a+b+c$. By AM-GM $p \geq 3$ and $q \geq 3$. So, $RHS \leq \frac{3}{2}$. By C-S $LHS = \sum \frac{(ab)^2}{ab+abc} \geq \frac{q^2}{q+3} \geq \frac{3}{2}$, because $2q^2-3q-9=(q-3)(2q+3) \geq 0$. Done!

Let $a,b,c$ be positive real numbers, that satisfy $abc=1$. Prove the inequality: $$\frac{ab}{1+c}+\frac{bc}{1+a}+\frac{ca}{1+b} \geq \frac{18}{(a+b+c)(1+a+b+c)}$$

nAalniaOMliO wrote: Let $a,b,c$ be positive real numbers, that satisfy $abc=1$. Prove the inequality: $$\frac{ab}{1+c}+\frac{bc}{1+a}+\frac{ca}{1+b} \geq \frac{27}{(a+b+c)(3+a+b+c)}$$

I got the same sol as @above.