Prove that $\frac{a}{b^2 + c}+\frac{b}{c^2 + a}+\frac{c}{a^2 + b} \ge \frac{3}{2}$ when $a+b+c=3$
$\sum_{cyc}\frac{a}{b+c^2}=\sum_{cyc}\frac{a^4}{a^3b+a^3c^2}\geq\frac{(a^2+b^2+c^2)^2}{\frac{1}{3}(a^2+b^2+c^2)^2+\frac{1}{3}(a^2+b^2+c^2)^2}=\frac{3}{2}$
(For numerator we use Titu's Lemma and simplify denominator :) )
Thanks to arqady
For more information vist https://artofproblemsolving.com/community/c6h401290p2234611