# 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

• You found that $x^4+y^4+z^2\ge 2\sqrt{2}xyz$, thus $\displaystyle \frac{x^4+y^4+z^2}{xyz}\ge \frac{2\sqrt{2}xyz}{xyz}=2\sqrt{2}$. – Galc127 Feb 18 '16 at 12:59
• Please fix the equation in title : $z^2$ and not $z^4$. I was lost ! Cheers. – Claude Leibovici Feb 18 '16 at 13:03

$$x^4+y^4+z^2=x^4+x^4+(\sqrt{2}x^2)^2=4x^4\\ xyz=xx\sqrt{2}x^2=\sqrt{2}x^4$$
We have, with AM-GM, that \begin{align} x^4+y^4+z^2&=4\cdot\frac{x^4+y^4+\tfrac12z^2+\tfrac12z^2}{4}\\ &\geq 4\sqrt[4]{x^4y^4z^4\cdot\tfrac14}\\ &=xyz\cdot 2\sqrt{2} \end{align} So that $$\frac{x^4+y^4+z^2}{xyz}\geq 2\sqrt{2}$$ With equality iff $x^4=y^4=\frac12z^2$, so for example $x=y=1$ and $z=\sqrt{2}$.