jmc

Subtracting $3$ from both sides gives $$\frac{x}{x-1}+\frac{x+2}{2x}-3\ge 0.$$Combining all the terms under a common denominator, we get $$\frac{x(2x)+(x+2)(x-1)-3(x-1)(2x)}{(x-1)(2x)}\ge 0,$$or $$\frac{-3x^2+7x-2}{2x(x-1)}\ge 0.$$Factoring the numerator, we get $$\frac{-(3x-1)(x-2)}{2x(x-1)}\ge 0.$$Making a sign table for the inequality $f(x)=\frac{(3x-1)(x-2)}{x(x-1)}\le 0,$ we get: \begin{array}{c|cccc|c} &$3x-1$ &$x-2$ &$x$ &$x-1$ &$f(x)$ \\ \hline$x<0$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+$\\ [.1cm]$0<x<\frac{1}{3}$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$-%%DISP_0%%amp;$-$\\ [.1cm]$\frac{1}{3}<x<1$ &$+%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$-%%DISP_0%%amp;$+$\\ [.1cm]$1<x<2$ &$+%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]$x>2$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]\end{array}Therefore, we have $f(x)<0$ when $0<x<\frac{1}{3}$ or $1<x<2.$ We also have $f(x)=0$ when $x=\frac{1}{3}$ or $x=2,$ so the whole solution set to the inequality is $$x\in \boxed{(0,\frac{1}{3}]\cup (1,2]}.$$