What's the maximum value?

Due to AM-GM: $\frac{xy^2z^2}{\sqrt[5]{16}}=\sqrt[5]{\frac{x^5y^{10}z^{10}}{2\cdot 2\cdot 2\cdot 2}}\leq \frac{x^5+\frac{y^5}{2}+\frac{y^5}{2}+\frac{z^5}{2}+\frac{z^5}{2}}{5}=\frac{x^5+y^5+z^5}{5} \\ \Rightarrow\; \frac{xy^2z^2}{x^5+y^5+z^5}\leq\frac{\sqrt[5]{16}}{5}=m$ Thus $(5m)^5 = (5\cdot \frac{\sqrt[5]{16}}{5})^5 = \boxed{16}$

Since x, y and z are positive,

$\frac{xy^{2}z^{2}}{x^{5}+y^{5}+z^{5}}$ is positive too.

So you can find the minimum value of multiplicative inverse of $\frac{xy^{2}z^{2}}{x^{5}+y^{5}+z^{5}}$ and the multiplicative inverse of the value will be the maximum value of $\frac{xy^{2}z^{2}}{x^{5}+y^{5}+z^{5}}$ .

$\frac { { x }^{ 5 }+{ y }^{ 5 }+{ z }^{ 5 } }{ x{ y }^{ 2 }{ z }^{ 2 } }$

$=\frac { { x }^{ 4 } }{ { y }^{ 2 }{ z }^{ 2 } } +\frac { { y }^{ 3 } }{ x{ z }^{ 2 } } +\frac { { z }^{ 3 } }{ x{ y }^{ 2 } }$

$=\frac { { x }^{ 4 } }{ { y }^{ 2 }{ z }^{ 2 } } +\frac { { y }^{ 3 } }{ 2x{ z }^{ 2 } } +\frac { { y }^{ 3 } }{ 2x{ z }^{ 2 } } +\frac { { z }^{ 3 } }{ 2x{ y }^{ 2 } } +\frac { { z }^{ 3 } }{ 2x{ y }^{ 2 } }$

$\ge 5\sqrt [ 5 ]{ \frac { { x }^{ 4 } }{ { y }^{ 2 }{ z }^{ 2 } } \frac { { y }^{ 3 } }{ 2x{ z }^{ 2 } } \frac { { y }^{ 3 } }{ 2x{ z }^{ 2 } } \frac { { z }^{ 3 } }{ 2x{ y }^{ 2 } } \frac { { z }^{ 3 } }{ 2x{ y }^{ 2 } } }$

$=\frac { 5 }{ \sqrt [ 5 ]{ 16 } }$

so the minimum value of $\frac { { x }^{ 5 }+{ y }^{ 5 }+{ z }^{ 5 } }{ x{ y }^{ 2 }{ z }^{ 2 } }$ = $\frac { 5 }{ \sqrt [ 5 ]{ 16 } }$

thus m= $\frac { \sqrt [ 5 ]{ 16 } }{ 5 }$ , $(5m)^{5}$ =16

First notice that $x$ cannot be zero.

So, divide the numerator and denominator by $x^{5}$ to get

$\large{\large{ \frac{ (\frac{y}{x})^{2} (\frac{z}{x})^{2} }{1 + (\frac{y}{x})^{5} + (\frac{z}{x})^{5} }}}$

now put $\frac{y}{x} = a , \frac{z}{x} = b$

by A.M - G.M inequality,

$1 + \frac{a^{5}}{2} + \frac{a^{5}}{2} + \frac{b^{5}}{2} + \frac{b^{5}}{2} \geq 5. \sqrt[5]{(\frac{a^{10} . b^{10}}{16})} =5.( \frac{a^2 . b^2 }{\sqrt[5]{16}})$

So,

$\frac{a^2 . b^2}{1 + a^5 + b^5} \leq (\frac{\sqrt[5]{16}}{5}). (\frac{a^2 . b^2}{a^2 . b^2}) = \frac{\sqrt[5]{16}}{5}$

Thus, our answer is $\boxed{16}$ .