I have the following inequality and I would like to get A on it's own:
$$ A \sqrt[q]{\frac{1}{1-q}} \left(\frac {-q}{1-q}\right) ≤ 1 $$
$q$ is just a parameter and $q$ root is like the square root/cube root only for any parameter.
I have tried this so far:
$$ A \sqrt[q]{\frac{1}{1-q}} ≤ \frac {1-q}{-q} $$
But I am unsure on what to do next. Can someone help please.
Encouraging you to take a look at this answer, I want to draw attention to the fact that you have to be careful about the signs when multiplying or dividing when dealing with inequalities.
If $\mathcal{Q} := \sqrt[q]{\frac{1}{1 - q}} \frac{-q}{1 - q} \ge 0$ (this is precisely the case when $q < 0$), you can manipulate your inequality by multiplying by its inverse $\mathcal{Q}^{-1} = \frac{1}{\sqrt[q]{\frac{1}{1 - q}}} \frac{1 - q}{-q}$: $$ A \le \frac{1}{\sqrt[q]{\frac{1}{1 - q}}} \frac{1 - q}{-q}. $$ If you were dealing with positive $q$, the inequality would be reversed by multiplying with the inverse, because the inverse is then negative as well: If $\mathcal{Q} < 0$, we have $$ A \ge \frac{1}{\sqrt[q]{\frac{1}{1 - q}}} \frac{1 - q}{-q}. $$
Edit: Also, the term $\mathcal{Q}$ can be simplified using $\sqrt[q]{x} = x^{\frac{1}{q}}$ as $$ \mathcal{Q} = \left( \frac{1}{q - 1}\right)^{\frac{1}{q + 1}} \cdot (-q) $$ and therefore its inverse can be stated in a simplified version: $$ \mathcal{Q}^{-1} = \frac{1}{-q} \cdot \left( \frac{1}{1 - q}\right)^{\frac{1}{1 - q}}. $$