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Let $N \in \mathbb{R}$ such that $N \geq 3$, $q > 1$ and $2^* := \frac{2N}{N - 2}$. Show that $\frac{1}{2^*} - \frac{2}{N(q-1) - 1} < 0 \iff q < 2^*$.

I would like to know how to prove the statement above. I tried to prove it, but I couldn't.

$\textbf{My attempt:}$

$(\Longrightarrow)$

$$\begin{align} \frac{1}{2^*} < \frac{2}{N(q-1) - 1} &\Longrightarrow \frac{N(q-1) - 1}{2} < 2^*\\ &\Longrightarrow q - 2 \overset{(N > 2)}{<} \frac{N(q-1) - N}{2} \overset{(N > 1)}{<} \frac{N(q-1) - 1}{2} < 2^* \end{align}$$

$(\Longleftarrow)$

$$\begin{align} 0 < q - 1 < q < 2^* &\Longrightarrow \frac{1}{2^*} < \frac{1}{q - 1}\\ &\Longrightarrow \frac{2}{2^*N} < \frac{2}{N(q-1)} < \frac{2}{N(q-1) - 1} \end{align}$$

George
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1 Answers1

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You want to show that $$\frac{N-2}{2N}-\frac{2}{N(q-1)-1}<0$$ if and only if $$q<\frac{2N}{N-2}$$ Is this correct? I would highly recommend never using a number as a variable, as it just looks incredibly confusing.

Anyways, lets just play around with the first equality and try to get the second one;

$$\frac{N-2}{2N}-\frac{2}{N(q-1)-1}<0$$ $$\frac{N-2}{2N}<\frac{2}{N(q-1)-1}$$ since $N-2$ is positive, we can get $$\frac{N(q-1)-1}{2}<\frac{2N}{N-2}$$ $$q<\frac{4}{N-2}+2$$ $$q<\frac{4+2(N-2)}{N-2}$$ $$q<\frac{2N}{N-2}$$

which is exactly what we want, except for one small problem, in step 2 we assumed both sides of the equation were positive. The only place where this is an issue is that $N(q-1)-1$ might be negative. I think it isn't too difficult to argue why we know $q$ cannot be negative, Although the final result should be $1<q<2^*$.

Moosh
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  • Thanks for the response! I forgot to put the hypothesis that $q > 1$. I have a question about your solution. Did you conclude that $q < \frac{4}{N - 2} + 2$ from $q < \frac{4}{N - 2} + \frac{1}{N} + 1 < \frac{4}{N - 2} + 2$ once that $N \geq 3$? Because from $\frac{N (q - 1) - 1}{2} < \frac{2N}{N - 2}$ you obtain that $q < \frac{4}{N - 2} + \frac{1}{N} + 1$. In that case, I can't do the work backwards to prove the reciprocal of the statement. Do you have an idea about how can I prove the reciprocal? – George Jun 15 '22 at 19:15
  • Every single step works the exact same way in both directions. Just keep track of where things need to be positive in order to not flip the inequality. – Moosh Jun 17 '22 at 14:59