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 $3\sqrt[3]{a^2b^2c^2}$?
Yet another way in which this can be shown using Schur's inequality in tandem with the AM-GM inequality is as follows:
$$
a^2+b^2+c^2+3(a^2b^2c^2)^{1/3}\geqslant a^{2/3}b^{4/3} + a^{4/3}b^{2/3} +b^{2/3}c^{4/3} + b^{4/3}c^{2/3} + a^{2/3}c^{4/3} + a^{4/3}c^{2/3} \\[2ex]= 2\left({a^{2/3}b^{4/3} + a^{4/3}b^{2/3}\over 2} + {b^{2/3}c^{4/3} + b^{4/3}c^{2/3}\over 2} + {a^{2/3}c^{4/3} + a^{4/3}c^{2/3}\over 2}\right)\\[2ex] \geqslant 2(ab + bc + ac)
$$
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:
$$\begin{align}
a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2}&amp;=6\sqrt[3]{a^2b^2c^2}+\delta_1\\
&amp;\ge6\sqrt[3]{a^2b^2c^2}+\delta_2\\
&amp;\ge2\left(3\sqrt[3]{a^2b^2c^2}+\frac{\delta_2}{2}\right)\\
&amp;\ge2\left(ab+bc+ca-\frac{\delta_2}{2}\right)\\
\end{align}$$
I am hoping this aproach triggers a thought in someones brain who can then come up with the final proof.