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 &gt; 3
\implies x-2 &gt; 1
\implies x(x-2) &gt; 3
\implies 1 &gt; \dfrac{3}{x(x-2)}
$
Your method is correct. You missed the sign in this step:
$$3-(x(x-2))&gt;0 \Rightarrow -x^2\color{red}{+}2x+3&gt;0 \Rightarrow (x+1)(x-3)&lt;0.$$
The solution is $x\in (-1,3)$, which contradicts the condition $x&gt;3$. So the original claim is true.

@fleablood's example: If $-1&lt;x&lt;3$, prove $1\le \frac{3}{x(x-2)}$.

Assume for $-1&lt;x&lt;3$, it is true: $1&gt;\frac{3}{x(x-2)}$.

Consider two cases:
$$\text{Case 1:} \ \ x\in (-1,1)\cup (2,3): \ \ \ 1&gt;\frac{3}{x(x-2)} \Rightarrow x(x-2)&gt;3 \Rightarrow x\in (-\infty,-1)\cup (3,+\infty) \Rightarrow \text{contradiction}; \\
\text{Case 2:} \ \ x\in (1,2): \ \ \ 1&gt;\frac{3}{x(x-2)} \Rightarrow x(x-2)&lt;3 \Rightarrow x\in (-1,3) \Rightarrow \text{not contradiction}.$$
Since Case 2 is not contradicting, the original claim is not true.