How to prove the inequality abc(a+b+c)2≤(a3+b3+c3)(ab+bc+ca)abc(a+b+c)2≤(a3+b3+c3)(ab+bc+ca)?

Question

I need to prove something like that:

For a,b,c>0$a,b,c>0$ prove: abc(a+b+c)2≤(a3+b3+c3)(ab+bc+ca)$abc(a+b+c{)}^2\le (a^3+b^3+c^3)(ab+bc+ca)$.

I know that 3abc≤(a3+b3+c3)$3abc\le (a^3+b^3+c^3)$, but then I derived 3(ab+bc+ca)≤(a+b+c)2$3(ab+bc+ca)\le (a+b+c{)}^2$, I can't move on.

Can anyone help me?

Answer 1

Hint: Using CS inequality,

(a3+b3+c3)(1a+1b+1c)⩾(a+b+c)2$$(a^3+b^3+c^3)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})⩾(a+b+c{)}^2$$

Answer 2

Hint:

−(a+b)(a+c)(b+c)(a2−ab−ac+b2−bc+c2)=abc(a+b+c)2−(a3+b3+c3)(ab+bc+ca)$$-(a+b)(a+c)(b+c)(a^2-ab-ac+b^2-bc+c^2)=abc(a+b+c{)}^2-(a^3+b^3+c^3)(ab+bc+ca)$$