<s>[INST] Prove $a^3+b^3+c^3\ge a^2+b^2+c^2$ if $ab+bc+ca\le 3abc$ 
if $a,b,c$ are positive real numbers and $ab+bc+ca\le 3abc$ Prove:$$a^3+b^3+c^3\ge a^2+b^2+c^2$$
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.

Things I have done so far: We are given $ab+bc+ca\le 3abc$ So$$2(ab+bc+ca)\le 6abc$$
so I can rewrite question as$$a^3+b^3+c^3+6abc \ge (a+b+c)^2$$
Also We can say$$9a^2b^2c^2 \ge (ab+bc+ca)^2 \ge 3abc(a+b+c)$$
My problem is is finding a way to use this facts and change the form of inequality to apply AM-GM.
 [/INST] First
$$3abc\ge ab+bc+ca\ge 3\sqrt[3]{a^2b^2c^2}.$$
So $abc\ge 1$.
It's then follows that $$a+b+c\ge 3\sqrt[3]{abc}\ge 3.$$
Now 
$$3(a^3+b^3+c^3)\ge (a^2+b^2+c^2)(a+b+c)$$
by Rearrangement Inequality or by
$$(a+b+c)(a^3+b^3+c^3)\ge (a^2+b^2+c^2)^2\ge (a^2+b^2+c^2)\frac{(a+b+c)^2}{3}.$$
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