Prove that $\frac{a}{b^2 + c}+\frac{b}{c^2 + a}+\frac{c}{a^2 + b} \ge \frac{3}{2}$ when $a+b+c=3$

Question

I've tried solving Pham Kim Hung's famous inequality problem which he posted on aops in April 2007 for 2 days now. This is the problem: Show that $$\frac{a}{b^2 + c}+\frac{b}{c^2 + a}+\frac{c}{a^2 + b} \ge \frac{3}{2} \tag{*}$$ when $a, b, c > 0$ and $a+b+c=3$.

I've tried with many theorems and tricks (like AM-GM, C-S, etc.) but haven't really figured out a way.

The closest I could get was by solving a similar problem like this.

So, if you know the proof of (*) please share here. I'd be much thankful to you.

Answer 1

$\sum_{cyc}\frac{a}{b+c^2}=\sum_{cyc}\frac{a^4}{a^3b+a^3c^2}\geq\frac{(a^2+b^2+c^2)^2}{\frac{1}{3}(a^2+b^2+c^2)^2+\frac{1}{3}(a^2+b^2+c^2)^2}=\frac{3}{2}$
(For numerator we use Titu's Lemma and simplify denominator :) )

Thanks to arqady
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