How to prove the inequality $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$?

Question

I need to prove something like that:

For $a,b,c>0$ prove: $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$.

I know that $3abc≤(a^3+b^3+c^3)$, but then I derived $3(ab+bc+ca) ≤ (a+b+c)^2$, I can't move on.

Can anyone help me?

Answer 1

Hint: Using CS inequality,

$$(a^3+b^3+c^3)\left(\frac1a+\frac1b+\frac1c\right)\geqslant (a+b+c)^2$$

Answer 2

Hint:

$$-(a+b)(a+c)(b+c)(a^2-ab-ac+b^2-bc+c^2)=abc(a+b+c)^2-(a^3+b^3+c^3)(ab+bc+ca)$$