115. Find the minimum value of the expression $\frac{(4+x)(1+x)}{x}$, where $x$ is a positive number.
Solution. Since

$$
\frac{(4+x)(1+x)}{x}=\frac{4}{x}+x+5=2\left(\frac{2}{x}+\frac{x}{2}\right)+5
$$

and $\frac{2}{x}+\frac{x}{2} \geqslant 2$ (for $c>0$, the inequality $c+\frac{1}{c} \geqslant 2$ holds, i.e., 2 is the minimum value of the sum of two reciprocals (see 627 from [2])), the expression $2\left(\frac{2}{x}+\frac{x}{2}\right)+5$ takes its minimum value, which is $9(2 \cdot 2+5=9)$.

Answer: 9.