Solving the inequality $\frac{x}{\sqrt{x+12}} - \frac{x-2}{\sqrt{x}} > 0$ I'm having troubles to solve the following inequality..
$$\frac{x}{\sqrt{x+12}} - \frac{x-2}{\sqrt{x}} > 0$$
I know that the result is $x>0$ and $x<4$ but I cannot find a way to the result..
Thanks in advance!

First write the inequality as
$$
{x\sqrt x-(x-2)\sqrt{x+12}\over \sqrt x\sqrt{x+12} }>0
$$
This holds if and only if $x>0$ and $x\sqrt x-(x-2)\sqrt{x+12}>0$.
We have
$$\eqalign{
&x\sqrt x-(x-2)\sqrt{x+12}>0\cr
\iff&x\sqrt x>(x-2)\sqrt{x+12}\cr 

}
$$
For $x>2$, squaring both sides of the above gives 
$$x^3>(x-2)^2 (x+12)$$
or
$$
0>4(2x^2-11x+12)
$$
or
$$
0>(2x-3)(x-4).
$$
The solution to the above is $3/2<x<4$. 
Thus for $x>2$, the only solutions to the original inequality are $2<x<4$.
For $0<x\le2$, we obviously have a solution to the original inequality (since both terms will be positive for $0<x<2$, and for $x=2$, we have the inequality ${2\over\sqrt{14}}>0$).
So the solution set is $(0,4)$.
