# Dealing with an inequality that involves a root on one side

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.

• What you've done so far is incorrect, or at least incomplete. You are assuming that ${-q\over1-q}\geq0$ Otherwise, the sense of the inequality will be reversed. What you've done is okay if you split the problem into two cases. Apr 28, 2019 at 22:57
• @saulspatz My lecturer recently did the following so I was trying to follow the same steps - A * (p/p+1)^p * (1/p+1) ≤ 1 which he solved and got A ≤ (p+1)(p+1/p)^p.
– May
Apr 28, 2019 at 23:03
• Was he assuming $p>0?$ An irrational power of a negative number isn't a real number, so he may well have been. But in your case, if say $0<q<1$ then the root makes sense, but $-q/(1q)<0$ so the multiplication would reverse the sign. Apr 28, 2019 at 23:10

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: $$\mathcal{Q}^{-1} = \frac{1}{-q} \cdot \left( \frac{1}{1 - q}\right)^{\frac{1}{1 - q}}.$$