<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] I think I have got close to proving this but I'm not sure if this is a valid proof - but here goes...
Using AM-GM we can show that:
$$a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\tag{1}$$
$$ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\tag{2}$$
We can also show that:
$$a^2+b^2+c^2\ge ab+bc+ca\tag{3}$$
We can therefore infer that:
$$a^2+b^2+c^2\ge ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\tag{4}$$
Using these facts we can say that:
$$a^2+b^2+c^2=3\sqrt[3]{a^2b^2c^2}+\delta_1\text{ where }\delta_1\ge0\tag{5}$$
$$ab+bc+ca=3\sqrt[3]{a^2b^2c^2}+\delta_2\text{ where }\delta_2\ge0\tag{6}$$
We can then use (4) to further infer that:
$$\delta_1\ge\delta_2\ge0$$
Finaly we can state that:
$$\begin{align}
a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2}&=6\sqrt[3]{a^2b^2c^2}+\delta_1\\
&\ge6\sqrt[3]{a^2b^2c^2}+\delta_2\\
&\ge2\left(3\sqrt[3]{a^2b^2c^2}+\frac{\delta_2}{2}\right)\\
&\ge2\left(ab+bc+ca-\frac{\delta_2}{2}\right)\\
\end{align}$$
I am hoping this aproach triggers a thought in someones brain who can then come up with the final proof.
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