Prove if $x\gt3$ then $1\ge\frac{3}{x(x-2)}$. I tried to prove it by contradiction.  
Suppose it is not true that  $1\ge\frac{3}{x(x-2)}$, so  $1\lt\frac{3}{x(x-2)}$.  Then  $\frac{3}{x(x-2)}-1\gt0$.  Multiply both sides of $\frac{3}{x(x-2)}-1\gt0$ by ${x(x-2)}$.
$(\frac{3}{x(x-2)}-1\gt0)({x(x-2)}\gt0(x(x-2)$
${3-(x(x-2)\gt0}$
${3-x^2-2x\gt0}$
${-x^2-2x+3\gt0}$
${-1(x^2+2x-3)\gt0}$
$-1\frac{(x-1)(x+3)}{-1}\gt0/-1$
${(x-1)(x+3)\lt0}$
At this point I really do not know what to do after this point or if I really even went about it the right way.  Thank you for the help.

$x > 3
\implies x-2 > 1
\implies x(x-2) > 3
\implies 1 > \dfrac{3}{x(x-2)}
$
