Positive integers x, y, z satisfy x y+z=160 . Compute the smallest...

$$\begin{aligned} & x+y z=\frac{160-z}{y}+y z \\ & \frac{160}{y}+\frac{z\left(y^2-1\right)}{y}=\frac{(160-z)}{y}+\frac{z y^2}{y}=\frac{160-z}{y}+z y \end{aligned}$$ At particle value of $z$ it is greater than equal to $2 \sqrt{z(160-z)}$ $\Rightarrow$ least value is $2 \sqrt{z(160-z)}$ but integer also Now least for least value $z$ is also Case I $z=1, x+y z=\frac{159}{y}+y$ $\Rightarrow$ minimum value is at $y=3$ which is 56 Case II $z=2, x+y z=\frac{158}{y}+2 y$ $\Rightarrow$ minimum value at $y=2$ which is 83 (rejected) Case III $z=3, x+y z=\frac{157}{y}+3 y$ $\Rightarrow$ minimum at $y=1$ which is 160 (rejected) Case IV $z=4, x+y z=\frac{156}{y}+4 y$ $\Rightarrow$ minimum at $y=6$ which is 50 (accept) Case V $z=5, x+y z=\frac{155}{y}+5 y$ minimum value at $y=5$ which is 56 (reject) Case VI $z=6, x+y z=\frac{154}{y}+6 y \geq 2 \sqrt{924}>50$ similarly in all the cases minimum value is greater than 50 . then answer is 50