(a) Using $abc=1$, and AM-HM inequality, \[ \sum \frac{ab}{a+b} = \sum \frac{1}{c(a+b)} \ge \frac{9}{2(ab+bc+ca)}. \](b) It suffices to show $a^2b + b^2c + c^2 a\ge ab+bc+ca$. To that end, let $a=x/y$, $b=y/z$ and $c=z/x$ for some $x,y,z>0$. After some algebra, it boils down verifying $x^3+y^3+z^3 \ge x^2y + y^2z + z^2x$. But this is immediate by AM-GM by noticing $(x^3+x^3+y^3)/3 \ge x^2y$ and applying analogous inequalities.

As in Bulgaria JBMO 2022 TST. In a), write $ab = 1/c$ etc. and finish by Titu's lemma. In b) we wish to show $a^2b + b^2c + c^2a \geq ab + bc + ca$ which follows by the AM-GMs $a^2b + c^2a + c^2a \geq 3\sqrt[3]{a^4bc^4} = 3ac$.