A basic inequality

Call the expression Q, it can be written as $Q=\frac{x^4}{x^2+x\sqrt{yz}}+\frac{y^4}{y^2+y\sqrt{xz}}+\frac{z^4}{z^2+z\sqrt{xy}}$ Now applying Titu's Lemma we get $Q\geq\frac{(x^2+y^2+z^2)^2}{x^2+y^2+z^2+\sqrt{xyz}(\sqrt{x}+\sqrt{y}+\sqrt{z})}$ We have this row of inequalities (by Cauchy-Schwarz): $3\sqrt{xyz}(\sqrt{x}+\sqrt{y}+\sqrt{z})\leq (\sqrt{xy}+\sqrt{yz}+\sqrt{zx})^2\leq (x+y+z)^2\leq 3(x^2+y^2+z^2)$ $\therefore Q\geq\frac{(x^2+y^2+z^2)^2}{x^2+y^2+z^2+\sqrt{xyz}(\sqrt{x}+\sqrt{y}+\sqrt{z})}\geq\frac{x^2+y^2+z^2}{2}\geq\frac{(x+y+z)^2}{6}\geq\frac{3}{2}$ The equality holds when $x=y=z=1$

$*Note$ : this problem can also be solved by AM-GM