<s>[INST] Prove that $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$ 
Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$

It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq ab+bc+ca+3\sqrt[3]{a^2b^2c^2}$. Then do I use the rearrangement inequality similarly on $3\sqrt[3]{a^2b^2c^2}$?
 [/INST] $x^3=a^2,y^3=b^2,z^3=c^2 \implies x^3+y^3+z^2 +3xyz \ge 2(\sqrt{(xy)^3}+\sqrt{(yz)^3}+\sqrt{(xz)^3})$
we have $x^3+y^3+z^3 +3xyz \ge xy(x+y)+yz(y+z)+xz(x+z)$
$xy(x+y)\ge 2xy\sqrt{xy}=2\sqrt{(xy)^3} \implies xy(x+y)+yz(y+z)+xz(x+z)\ge 2(\sqrt{(xy)^3}+\sqrt{(yz)^3}+\sqrt{(xz)^3})$
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