5. Given that $x, y, z$ are 3 real numbers greater than or equal to 1, then
$$
\left(\frac{\sqrt{x^{2}(y-1)^{2}+y^{2}}}{x y}+\frac{\sqrt{y^{2}(z-1)^{2}+z^{2}}}{y z}+\frac{\sqrt{z^{2}(x-1)^{2}+x^{2}}}{z x}\right)^{2}
$$

the sum of the numerator and denominator of the minimum value written as a simplified fraction is $\qquad$ .
Solve:
$$
\frac{\sqrt{x^{2}(y-1)^{2}+y^{2}}}{x y}=\sqrt{\left(1-\frac{1}{y}\right)^{2}+\frac{1}{x^{2}}} \leqslant \frac{\sqrt{2}}{2}\left[\left(1-\frac{1}{y}\right)+\frac{1}{x}\right] \text {, }
$$

Similarly, we get $\frac{\sqrt{y^{2}(z-1)^{2}+z^{2}}}{y z} \leqslant \frac{\sqrt{2}}{2}\left[\left(1-\frac{1}{z}\right)+\frac{1}{y}\right], \frac{\sqrt{z^{2}(x-1)^{2}+x^{2}}}{z x} \leqslant \frac{\sqrt{2}}{2}\left[\left(1-\frac{1}{x}\right)+\frac{1}{z}\right]$,
Adding the three inequalities, we get $\left(\frac{\sqrt{x^{2}(y-1)^{2}+y^{2}}}{x y}+\frac{\sqrt{y^{2}(z-1)^{2}+z^{2}}}{y z}+\frac{\sqrt{z^{2}(x-1)^{2}+x^{2}}}{z x}\right)^{2} \leqslant\left(\frac{3 \sqrt{2}}{2}\right)^{2}=\frac{9}{2}$,
Therefore, the sum of the numerator and denominator of the simplest fraction of the original expression's minimum value is 11.