Inequality problem with 4th root. $ \frac{6x}{x-2} - \sqrt{\frac{12x}{x-2}} - 2\sqrt[4]{\frac{12x}{x-2}}>0 $
I've tried putting $t= \frac{6x}{x-2} $ and play algebraically, using square of sum, but still no luck, any help?
 A: Instead, let's put $$t = \sqrt[4]{\frac{12x}{x-2}}.$$
Then:
$$\frac{6x}{x-2} - \sqrt{\frac{12x}{x-2}} - 2\sqrt[4]{\frac{12x}{x-2}}>0 \Rightarrow \\
\frac{1}{2}t^4 - t^2 - 2t>0 \Rightarrow \\
t^4 - 2t^2 - 4t > 0 \Rightarrow \\
t(t-2)(t^2+2t+2) >0.
$$
Then:
$$t <0 ~ \vee t > 2, $$
or equivalently:
$$\sqrt[4]{\frac{12x}{x-2}} <0 ~ \vee \sqrt[4]{\frac{12x}{x-2}} > 2.$$
For sure, $\sqrt[4]{\frac{12x}{x-2}} >0$, then we deal only with $\sqrt[4]{\frac{12x}{x-2}} > 2$. 
Furthermore, the argument of the $4$th root must be positive. That is:
$$\frac{12x}{x-2} > 0 \Rightarrow x < 0 \vee x > 2.$$
Regarding $\sqrt[4]{\frac{12x}{x-2}} > 2$, we have that:
$$\frac{12x}{x-2} > 2^4.$$
We have two cases:


*

*$x < 0$. In this case we get that $x > 8$. But this is a contradiction.

*$x > 2$. In this case we get that $x < 8$.


Finally, we found that:
$$ 2 < x < 8.$$
A: Let $\sqrt[4]{\frac{12x}{x-2}}=t$. Hence, $t\geq0$ and 
$$\frac{t^4}{2}-t^2-2t>0$$ or
$$t(t^3-2t-4)>0$$ or
$$t(t^3-2t^2+2t^2-4t+2t-2)>0$$ or
$$t(t-2)(t^2+2t+2)>0$$ or
$$t(t-2)((t+1)^2+1)>0$$ or 
$$t(t-2)>0$$ and since $t\geq0$, we obtain $t>2$ or
$$\frac{12x}{x-2}>16$$ or
$$\frac{3x}{x-2}-4>0$$ or
$$\frac{8-x}{x-2}>0$$  or$$2<x<8$$
Done!
