Inequality problem with 4th root.

Question

$ \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?

Answer 1

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:

1. $x < 0$. In this case we get that $x > 8$. But this is a contradiction.
2. $x > 2$. In this case we get that $x < 8$.

Finally, we found that:

$$ 2 < x < 8.$$

Answer 2

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