Well, it looks symmetric!

We see that $\dfrac{(x^4+1)(y^4+1)(z^4+1)}{xy^2z}=\left(x^3+\dfrac{1}{3x}+\dfrac{1}{3x}+\dfrac{1}{3x}\right)\left(y^2+\dfrac{1}{y^2}\right)\left(z^3+\dfrac{1}{3z}+\dfrac{1}{3z}+\dfrac{1}{3z}\right)$

Using AM-GM on each of these pieces gives $\left(x^3+\dfrac{1}{3x}+\dfrac{1}{3x}+\dfrac{1}{3x}\right)\left(y^2+\dfrac{1}{y^2}\right)\left(z^3+\dfrac{1}{3z}+\dfrac{1}{3z}+\dfrac{1}{3z}\right)\ge \left(4\sqrt[4]{\dfrac{1}{27}}\right)(2)\left(4\sqrt[4]{\dfrac{1}{27}}\right)=\dfrac{32\sqrt{3}}{9}$

so our answer is $32+3+9=\boxed{44}$ .

Equality case is when $(x,y,z)=\left(\dfrac{1}{\sqrt[4]{3}},1,\dfrac{1}{\sqrt[4]{3}}\right)$

Note that by AM-GM inequality, $y^4+1\geq 2y^2$ so we can cancel the $y^2$ . This trick doesn't work for others. So we split up the $1$ so that we have $4$ terms to cancel the $4$ th power. $x^4 + 1 = x^4 + \dfrac{1}{3} +\dfrac{1}{3}+\dfrac{1}{3} \geq 4\sqrt[4]{\dfrac{x^4}{3^3}} = \dfrac{4x}{\sqrt[4]{3^3}}$ $z^4+1=z^4+\dfrac 1 3+\dfrac 1 3+\dfrac 1 3\geq 4\sqrt[4]{\dfrac{z^4}{3^3}}=\dfrac{4z}{\sqrt[4]{3^3}}$

Therefore: $\dfrac{(x^4+1)(y^4+1)(z^4+1)}{xy^2z}\geq \dfrac{4x}{\sqrt[4]{3^3}}\cdot 2y^2\cdot \dfrac{4z}{\sqrt[4]{3^3}}\cdot \dfrac{1}{xy^2 z}=\dfrac{32\sqrt{3}}{9}$

Thus the answer is $\fbox{44}$ .

Note: the equality occurs when $y^4=1\implies y=1$ , $x^4=z^4=1/3$ $\implies x=z=1/\sqrt[4]{3}$ .

We see that $\frac { \partial }{ \partial x } \left( \frac { ({ x }^{ 4 }+1)({ y }^{ 4 }+1)({ z }^{ 4 }+1) }{ x{ y }^{ 2 }z } \right) = 0$ for $x = 3^{1/4}$ and $\frac { \partial }{ \partial y } \left( \frac { ({ x }^{ 4 }+1)({ y }^{ 4 }+1)({ z }^{ 4 }+1) }{ x{ y }^{ 2 }z } \right) = 0$ for $y=1$ . The equation is symmetric in $x$ and $z$ , and the functions are increasing in $x, y$ and $z$ . Hence, the minimum value is $\frac{32\sqrt3}{9}$ . Which gives us the answer $\boxed{44}$