<s>[INST] 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!
 [/INST] We can combine the two fractions on the left-hand side to get
$$\frac{x\sqrt x-(x-2)\sqrt{x+12}}{\sqrt{x+12}\sqrt{x}}>0$$
and since we need for the denominator to be defined and not $0$, this gives us $x>0$. We can now multiply both sides by $\sqrt{x+12}\sqrt{x}$ to get 
$$x\sqrt x-(x-2)\sqrt{x+12}>0$$
which we rewrite as 
$x\sqrt x>(x-2)\sqrt{x+12}$ and square both sides giving us
$$x^3>\pm(x-2)^2(x+12)=\pm(x^3+8x^2-44x+48)$$
where the $\pm$ is determined by the sign of $x-2$, since if $x\geq 2$ then squaring does not change the sign of either side while if $0<x<2$ then it does.
When $x\geq 2$ this simplifies to $0>8x^2-44x+24$. Applying the quadratic equation, we get that $0=8x^2-44x+48$ at $x=\frac{3}{2},4$ and we can see that between these two values the inequality holds, so we know that our desired inequality holds for $x\geq 2$. When $0<x<2$ this simplifies to $2x^3+8x^2-44x+48>0$, which is true for all positive $x$, so the desired inequality holds for $0<x<2$. Thus it holds in the region $0<x<4$, but not outside.
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