contest math - 3 variable symmetric inequality

Show that for positive reals $a,b,c$,
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{3a^3+3b^3+3c^3}{2a^2+2b^2+2c^2}$

What I did was WLOG $a+b+c=1$ (since the inequality is homogenous)
Then I substituted into the LHS to get $\sum_{\text{cyc}} \frac{a^2}{1-a}\geq \frac{3a^3+3b^3+3c^3}{2a^2+2b^2+2c^2}$. Now I'm not sure if I should clear the denominators and expand and try to use Muirhead+Schur? (Clearing the denominators seems quite tedious)?

Any ideas are appreciated.

Answer

your inequalitiy is equivalent to
$$\left(a^3-2 a b c+b^3+c^3\right) \left(2 a^3-a^2 b-a^2 c-a b^2-a c^2+2 b^3-b^2
c-b c^2+2 c^3\right)\geq 0$$
since $$a^3+b^3+c^3\geq 3abc>2abc$$ is the first factor positive,
and
$$2(a^3+b^3+c^3)\geq ab(a+b)+ac(a+c)+bc(b+c)$$
since $a^3+b^3=(a+b)(a^2+b^2-ab)\geq ab(a+b)$ etc is the second factor also positive.