jmc

Moving all the terms to the left-hand side, we have $$\frac{1}{r}-\frac{1}{r-1}-\frac{1}{r-4}>0.$$To solve this inequality, we find a common denominator: $$\frac{(r-1)(r-4)-r(r-4)-r(r-1)}{r(r-1)(r-4)}>0,$$which simplifies to $$\frac{-(r-2)(r+2)}{r(r-1)(r-4)}>0.$$Therefore, we want the values of $r$ such that $$f(r)=\frac{(r-2)(r+2)}{r(r-1)(r-4)}<0.$$To solve this inequality, we make the following sign table: \begin{array}{c|ccccc|c} &$r-2$ &$r+2$ &$r$ &$r-1$ &$r-4$ &$f(r)$ \\ \hline$r<-2$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-$\\ [.1cm]$-2<r<0$ &$-%%DISP_0%%amp;$+%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+$\\ [.1cm]$0<r<1$ &$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-$\\ [.1cm]$1<r<2$ &$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$-%%DISP_0%%amp;$+$\\ [.1cm]$2<r<4$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$-%%DISP_0%%amp;$-$\\ [.1cm]$r>4$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]\end{array}Putting it all together, the solutions to the inequality are $$r\in \boxed{(-\infty ,-2)\cup (0,1)\cup (2,4)}.$$