Prove that $\sum_\circ\frac{a^3+3b^3}{5a+b}\ge\frac23(a^2+b^2+c^2)$ for $a,b,c>0$ 
Prove the following for a, b and c being positive real numbers :-
$\sf{{ \dfrac{ a^3 + 3b^3 }{5a+b} + \dfrac{ b^3 + 3c^3}{5b+c} + \dfrac{ c^3 + 3a^3}{5c+a} \geqslant \dfrac{2}{3} (a^2 + b^2 + c^2) }} $

So here we have a question in which we have to prove that the equation which we are provided with, has real values of a, b and c.
I know on this site we have to show our work while asking a question, but actually I'm unable to even start it. Please can you give me hint for solving this question? Or at  least hint where to start.
I got this question while practicing some previous year examination papers for my future test which is next month.

Split our the fractions; use CS to grouped summands; use $ab+bc+ca \le a^2+b^2+c^2$:
$$\begin{align}
&\dfrac{ a^3 + 3b^3 }{5a+b} + \dfrac{ b^3 + 3c^3}{5b+c} + \dfrac{ c^3 + 3a^3}{5c+a} \\
&= \color{blue}{\dfrac{ a^4 }{5a+b}} + 3\color{fuchsia}{\frac{b^3}{5a+b}} \\
&+ \color{blue}{\dfrac{ b^3}{5b+c}} + 3\color{fuchsia}{\frac{c^3}{5b+c}} \\
&+ \color{blue}{\dfrac{ c^3}{5c+a}} + 3\color{fuchsia}{\frac{a^3}{5c + a}} \\
&= \color{blue}{\dfrac{ a^4 }{5a^2+ab} + \dfrac{ b^4}{5b^2+bc} + \dfrac{ c^4}{5c^2+ac}} \\
&+ 3\left(\color{fuchsia}{\frac{b^4}{5ab+b^2} + \frac{c^4}{5bc+c^2} + \frac{a^4}{5ca + a^2}}\right) \\
&\ge \frac{(a^2+b^2+c^2)^2}{5(a^2+b^2+c^2) + ab+bc+ca} + 3\frac{(a^2+b^2+c^2)^2}{a^2+b^2+c^2 + 5(ab+bc+ca)} \\
&\ge \frac{(a^2+b^2+c^2)^2}{5(a^2+b^2+c^2) + a^2+b^2+c^2} + 3\frac{(a^2+b^2+c^2)^2}{a^2+b^2+c^2 + 5(a^2+b^2+c^2)}  \\
&= \left(\frac{1}{6} + \frac{3}{6}\right)(a^2+b^2+c^2)\\
&= \dfrac{2}{3} (a^2 + b^2 + c^2)\end{align}$$
