Prove inequality $\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\geq(a-b)^2+(b-c)^2+(c-a)^2 $ I cannot prove the following inequality.

Let $a,b,c$ be positive numbers.Prove that:
  $\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}\geq(a-b)^2+(b-c)^2+(c-a)^2. $

I tried to use Cauchy inequality. I think the functions $f(a,b,c)=\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}$ and $g(a,b,c)=(a-b)^2+(b-c)^2+(c-a)^2$ are  symmetric with respect to $a,b,c.$  

$$\begin{align}&\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b} - (a-b)^2-(b-c)^2-(c-a)^2 \\ =& \sum\limits_{cyc} \left(\frac{a^3}{b+c} - 2a^2 +a(b+c)\right) \\ =& \sum\limits_{cyc} \frac{a(a-b-c)^2}{b+c} \ge 0 \end{align}$$
