We need to prove that $\sum_{cyc}x\sqrt{(x^2+xy+y^2)(x^2+xz+z^2)}\geq\sqrt{3\prod_{cyc}(x^2+xy+y^2)}.$ Now, by C-S $(x^2+xy+y^2)(x^2+xz+z^2)=\left(\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2\right)\left(\left(x+\frac{z}{2}\right)^2+\frac{3}{4}z^2\right)\geq$ $\geq\left(\left(x+\frac{y}{2}\right)\left(x+\frac{z}{2}\right)+\frac{3}{4}yz\right)^2=\left(x^2+\frac{x(y+z)}{2}+yz\right)^2.$ Thus, it remains to prove that $\left(\sum_{cyc}(2x^3+x^2y+x^2z+2xyz)\right)^2\geq12\prod_{cyc}(x^2+xy+y^2).$ Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Since $\prod\limits_{cyc}(x^2+xy+y^2)$ is a linear expression of $w^3$, we see that $\left(\sum_{cyc}(2x^3+x^2y+x^2z+2xyz)\right)^2-12\prod_{cyc}(x^2+xy+y^2)=81w^6+A(u,v^2)w^3+B(u,v^2),$ which says that the inequality $\left(\sum_{cyc}(2x^3+x^2y+x^2z+2xyz)\right)^2-12\prod_{cyc}(x^2+xy+y^2)\geq\left(\sum_{cyc}(x^3-x^2y-x^2z+xyz)\right)^2$ is a linear inequality of $w^3$, which says that it remains to prove the last inequality for an extremal value of $w^3$, which happens in the following cases.
- $w^3\rightarrow0^+$.
Let $z\rightarrow0^+$ and $y=1$.
We obtain $(x-1)^2(3x^4+14x^3+12x^2+12x+4)\geq0,$ which is obvious;
- $y=z=1$, which gives $(x-1)^2(x^3+6x^2+9x+8)x\geq0.$ Done!