From AM-GM we have $x^3+x\geq 2x^2$, $y^3+y\geq 2y^2$, $z^3+z\geq 2z^2$, so it is enough to prove that $$x+y+z\geq \sqrt{3(x+y+z)}$$which is equivalent to $x+y+z\geq 3$. It follows from $x+y+z\geq\sqrt{xy} + \sqrt{yz} + \sqrt{zx}$. Equality is attained when $x=y=z=1$.

Cobra23 wrote: From AM-GM we have $x^3+x\geq 2x^2$, $y^3+y\geq 2y^2$, $z^3+z\geq 2z^2$, so it is enough to prove that $$x+y+z\geq \sqrt{3(x+y+z)}$$which is equivalent to $x+y+z\geq 3$. It follows from $x+y+z\geq\sqrt{xy} + \sqrt{yz} + \sqrt{zx}$. Equality is attained when $x=y=z=1$. Very nice.

Let $x=a^2,y=b^2,z=c^2$ to avoid square roots We know that $ab+bc+ca=3$ By AM-GM, $LHS\geq\sqrt{2}(a^2+b^2+c^2)$ After squaring, it suffices to prove $a^2+b^2+c^2\geq 3$, which is true, as $a^2+b^2+c^2\geq ab+bc+ca$

Let $\sqrt{x}=a,\sqrt{y}=b$ and $\sqrt{z}=c$ so we have that $ab+bc+ca=3$ And we should prove; $\sum \sqrt{a^6+a^2} \ge \sqrt{6\sum a^2}$ Notice that $LHS=\sum a\sqrt{a^4+1}$ By AM-GM $\sqrt{a^4+1}\ge a\sqrt{2}$ so $a\sqrt{a^4+1}\ge a^2\sqrt{2}$ $\sum a\sqrt{a^4+1} \ge \sum a^2\sqrt{2}$ $(a^2+b^2+c^2)\sqrt{2}\ge \sqrt{6(a^2+b^2+c^2)}$ $\to \sqrt{a^2+b^2+c^2}\ge \sqrt{3}\to a^2+b^2+c^2\ge 3$ which is true because $a^2+b^2+c^2\ge ab+bc+ca=3$