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 $3\sqrt[3]{a^2b^2c^2}$?

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality it's $f(v^2)\geq0$, where $f(v^2)=3u^2-4v^2+w^2$.
Thus, $f$ is a linear function, which says that $f$ get's a minimal value for an extremal value of $v^2$, 
which happens for equality case of two variables.
Let $b=a=x^3$ and $c=1$.
Hence, we need to prove that $x^6+2+3x^2\geq2(2x^3+1)$,
which is $(x-1)^2(x^2+2x+3)x^2\geq0$. Done!
