Another way.

The first step is a homogenization as Fedor Petrov.

We need to prove that: $$x^2+y^2+z^2+3\sqrt[3]{x^2y^2z^2}\geq2(xy+xz+yz).$$ Now, for $xyz=0$ our inequality is obvious.

let $xyz>0$, $x=e^{\frac{a}{2}}$,$y=e^{\frac{b}{2}}$ and $z=e^{\frac{c}{2}}.$

Thus, we need to prove that $$\sum_{cyc}e^a+3e^{\frac{a+b+c}{3}}\geq2\sum_{cyc}e^{\frac{a+b}{2}},$$ which is the Popoviciu's inequality for the convex function $f(x)=e^x$.

About Popoviciu see here: https://en.wikipedia.org/wiki/Popoviciu%27s_inequality

Another way.

The first step is a homogenization as Fedor Petrov.

We need to prove that: $$x^2+y^2+z^2+3\sqrt[3]{x^2y^2z^2}\geq2(xy+xz+yz).$$ Now, for $xyz=0$ our inequality is obvious.

let $xyz>0$, $x=e^{\frac{a}{2}}$,$y=e^{\frac{b}{2}}$ and $z=e^{\frac{c}{2}}.$

Thus, we need to prove that $$\sum_{cyc}e^a+3e^{\frac{a+b+c}{3}}\geq2\sum_{cyc}e^{\frac{a+b}{2}},$$ which is the T.Popoviciu's inequality for the convex function $f(x)=e^x$.

About Popoviciu see here: https://en.wikipedia.org/wiki/Popoviciu%27s_inequality

Also, the inequality $$x^2+y^2+z^2+3\sqrt[3]{x^2y^2z^2}\geq2(xy+xz+yz)$$ we can get for $n=3$ from the following F.Shleifer's inequality.

Let $x_i\geq0$. Prove that: $$(n-1)\sum_{i=1}^nx_i^2+n\sqrt[n]{\prod_{i=1}^nx_i^2}\geq\left(\sum_{i=1}^nx_i\right)^2.$$