<s>[INST] If $x,y,z$ are positive real number number, Then minimum value of $\frac{x^4+y^4+z^2}{xyz}$ 
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 
 [/INST] $$x^4+y^4+z^2=x^4+x^4+(\sqrt{2}x^2)^2=4x^4\\
xyz=xx\sqrt{2}x^2=\sqrt{2}x^4$$
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