<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] We only need to show that
$$(bc+ca+ab)(a^3+b^3+c^3)\ge3abc(a^2+b^2+c^2)$$
which is equivalent to
$$\sum_\text{cyc}a(b^4+c^4-b^3c-bc^3)\ge0$$
or
$$\sum_\text{cyc}a(b-c)^2(b^2+bc+c^2)\ge0$$
where $\sum_\text{cyc}$ is a summing operator $\sum_\text{cyc}f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b)$.
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