Inequality. Prove that $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}+2\sqrt{3abc(a+b+c)}\geq3(a^2+b^2+c^2)$ Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}+2\sqrt{3abc(a+b+c)}\geq3(a^2+b^2+c^2)$$
I tried SOS, C-S, the $uvw$'s technique and more, but without success. 

Let $p = a+b+c, \ q = ab+bc+ca, \ r = abc$. We have $q^2 \ge 3pr$.
Using AM-GM, we have $\sqrt{3pr} \ge \frac{6pqr}{q^2 + 3pr}$. Thus, it suffices to prove that
$$\frac{a^3}{b} + \frac{b^3}{c} + \frac{c^3}{a} + 
\frac{12(a+b+c)(ab+bc+ca)abc}{(ab+bc+ca)^2 + 3(a+b+c)abc} \ge 3(a^2+b^2+c^2).$$
I can solve it by the Buffalo Way. Maybe someone can give nice proofs.
WLOG, assume that $c = \min(a,b,c)$. Let $b = c+ s, \ a = c+t; \ s,t\ge 0$. 
We need to prove that $q_7c^7 + q_6c^6 + q_5c^5 + q_4c^4 + q_3c^3 + q_2c^2 + q_1c + q_0 \ge 0$ where
\begin{align}
q_7 &= 12 s^2-12 s t+12 t^2, \\
q_6 &= 32 s^3+48 s^2 t-60 s t^2+32 t^3, \\
q_5 &= 45 s^4+114 s^3 t+3 s^2 t^2-102 s t^3+45 t^4, \\
q_4 &= 31 s^5+113 s^4 t+115 s^3 t^2-113 s^2 t^3-43 s t^4+31 t^5, \\
q_3 &= 7 s^6+67 s^5 t+91 s^4 t^2+17 s^3 t^3-91 s^2 t^4+15 s t^5+7 t^6, \\
q_2 &= 14 s^6 t+44 s^5 t^2+23 s^4 t^3-18 s^3 t^4-13 s^2 t^5+7 s t^6, \\
q_1 &= 8 s^6 t^2+8 s^5 t^3-3 s^3 t^5+s^2 t^6, \\
q_0 &= s^6 t^3.
\end{align}
It is not hard to prove that $q_7, q_6, \cdots, q_0\ge 0$. Omitted. We are done.
