Minimum Value of x^4 + y^4 + z^2 / xyz for Positive Reals

If $x,y,z$ are positive real number number, Then minimum value of $\displaystyle \frac{x^4+y^4+z^2}{xyz}$

$\bf{My\; Try::}$ Given $x,y,z>0.$ So Using $\bf{A.M\geq G.M\;,}$ We get

$$\displaystyle x^4+y^4\geq 2x^2y^2$$ and then Using $$2x^2y^2+z^2\geq 2\sqrt{2}|xyz|\geq 2\sqrt{2}xyz$$

So we get $$x^4+y^4+z^4\geq 2x^2y^2+z^2 \geq 2\sqrt{2}xyz$$

and equality hold when $x^4=y^4$ and $2x^2y^2=z^2$

So we get $x=y$ and $z=\sqrt{2}x^2$, Now Minimum occur at $\left(x,x,\sqrt{2}z^2\right)$

So at that point we get $\displaystyle \left[\frac{x^4+y^4+z^4}{xyz}\right]_{\bf{\min}} = \sqrt{2}+2x^2$

and Answer given is $2\sqrt{2}$, I did not understand how this can happen

olz explain me, Thanks

My Question is