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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.

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    $\begingroup$ 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. $\endgroup$
    – saulspatz
    Apr 28, 2019 at 22:57
  • $\begingroup$ @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. $\endgroup$
    – May
    Apr 28, 2019 at 23:03
  • $\begingroup$ 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. $\endgroup$
    – saulspatz
    Apr 28, 2019 at 23:10

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

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  • $\begingroup$ @May, if the answer was of help, please vote it up and/or mark it as the answer. $\endgroup$
    – NoChance
    Apr 28, 2019 at 23:19
  • $\begingroup$ @NoChance I did but it gave the notification 'Votes cast by those with less than 15 reputation are recorded, but do not change the publicly displayed post score.' $\endgroup$
    – May
    Apr 28, 2019 at 23:20
  • $\begingroup$ @May, maybe you got to wait :) $\endgroup$
    – NoChance
    Apr 28, 2019 at 23:23

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