How prove this inequality $ \frac{b^3+c^3}{a}+\frac{c^3+a^3}{b}+\frac{a^3+b^3}{c} \ge 2(a^2+b^2+c^2)+3\left((b-c)^2+(c-a)^2+(a-b)^2\right)$ let $a,b,c$ are positive numbers, show that

$$ \frac{b^3+c^3}{a}+\frac{c^3+a^3}{b}+\frac{a^3+b^3}{c} \ge 2(a^2+b^2+c^2)+3\left((b-c)^2+(c-a)^2+(a-b)^2\right)\cdots (1)$$

my try:

$$\Longleftrightarrow \frac{b^3+c^3}{a}+a^2+\frac{c^3+a^3}{b}+b^2+\frac{a^3+b^3}{c}+c^2 \ge 3(a^2+b^2+c^2)+3\left((b-c)^2+(c-a)^2+(a-b)^2\right)$$
  $$\left(a^3+b^3+c^3\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge 9(a^2+b^2+c^2)-6(ab+bc+ac)$$
  $$(a^3+b^3+c^3)(ab+bc+ac)\ge abc[9(a^2+b^2+c^2)-6(ab+bc+ac)]$$

some days ago,I have ask this same problem：How prove this inequality $(a^3+b^3+c^3)(ab+bc+ac)\ge 6abc(a^2+b^2+c^2-ab-bc-ac)$
then I can't prove it,Thank you 
maybe $(1)$ have other nice methods?

Hint: Read Arash's proof and tighten up his loose bounds.

Similar to the proof by Arash Beh in your other linked question, we have
$$(a^3 + b^3 + c^3) = (a+b+c) ( a^2 + b^2 +c^2 - ab - bc -ca)  + 3abc $$
He showed that
$$ (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)(ab+bc+ca) \geq 9abc( a^2 + b^2 + c^2 - ab - bc -ca),$$
which is obvious because 
$$(a+b+c) (ab+bc+ca) \geq 9abc.$$
We then add
$$ 3abc (ab+bc+ca) = 3 abc( ab+bc+ca)$$
to both sides, and we are done.
