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

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?

$$(a^3+b^3+c^3)\left(\frac1a+\frac1b+\frac1c\right)\geqslant (a+b+c)^2$$
$$-(a+b)(a+c)(b+c)(a^2-ab-ac+b^2-bc+c^2)=abc(a+b+c)^2-(a^3+b^3+c^3)(ab+bc+ca)$$