If $a^2+b^2+c^2=3$ so $a^2b+b^2c+c^2a\geq3\sqrt[3]{a^2b^2c^2}$ Let $a$, $b$ and $c$ be non-negative numbers such that $a^2+b^2+c^2=3$. Prove that:
$$a^2b+b^2c+c^2a\geq3\sqrt[3]{a^2b^2c^2}$$
I tried Rearrangement, uvw, BW and more, but without any success.

$$\Leftrightarrow (a^2b+b^2c+c^2a)^2 \ge 9(abc)^{4/3}=3(abc)^{4/3}(a^2+b^2+c^2)$$
$$\sum_{cyc}a^4b^2+2\sum_{cyc}a^3bc^2 \ge 3\sum_{cyc}\sqrt[3]{a^{10}b^4c^4}$$
By AM-GM : 
$$a^4b^2+2a^3bc^2 \ge 3\sqrt[3]{a^{10}b^4c^4}$$
