20. The function $f(x)$ satisfies $f\left(\frac{x}{y}\right)=f(x)-f(y)$, and $f(x)$ is not identically zero. When $x>1$, $f(x)>0$. Solve the inequality: $f(x)+f\left(x+\frac{1}{2}\right)<0$.
20. Given $f\left(\frac{x}{y}\right)=f(x)-f(y)$, then $f(1)=0, f\left(\frac{1}{y}\right)=-f(y), f(x y)=f(x)+f(y)$, the original inequality is transformed into $f\left[x\left(x+\frac{1}{2}\right)\right]1, f(x)>0$, which can prove that $f(x)$ is increasing on $(0,+\infty)$. Therefore, the inequality is transformed into
$$
\left\{\begin{array}{l}
x\left(x+\frac{1}{2}\right) \neq 0 . \\
\left|x\left(x+\frac{1}{2}\right)\right|<1 .
\end{array}\right.
$$

Solving this, we get $\frac{1-\sqrt{17}}{4}<x<0$ or $0<x<\frac{1}{2}$ or $\frac{1}{2}<x<\frac{1+\sqrt{17}}{4}$.