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Let

$$f(b) = \frac{(r t)^{1/b}-1}{r^{1/b}-1} $$

Show that $f'(b)>0$ for all $b\geq1$ as long as $0<t<1$, and $r>1/t$, or provide a counter-example.


My solution attempt:

$$f'(b)=\frac{r^{1/b} \log (r) \left((r t)^{1/b}-1\right)-\left(r^{1/b}-1\right) (r t)^{1/b} \log (r t)}{b^2 \left(r^{1/b}-1\right)^2} $$

Noting that the denominator is always positive and that I can factor out an $r^{1/b}$ of the numerator (which is also positive). The problem reduces to needing to show that

$((rt)^{1/b}-1)\log(r) - t^{1/b} (r^{1/b}-1)\log(rt) > 0$

if $0<t<1$, $r>1/t$, and $b\geq1$

I've tried much algebraic manipulation of the left-hand side, but haven't been able to come up with something yet. I've tried plotting this expression as a function of $b$ for various $r$ and $t$ that satisfy the assumptions and it seems to have held for every combination I have tried. So far in the expression above I have not used the fact that $r>1/t$ and $0<t<1$, which I know is crucial, as it's easy to show this expression is less than zero for several cases where $r<1/t$. If it turns out the statement is false, do any other conditions make it true?

EDIT: note I fixed the typo pointed out in a previous version of John Omielan's answer.

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  • $\begingroup$ Note I updated my answer to explain that $f'(b) \gt 0$ is actually true more generally for all $b \gt 0$. I'm not sure why the question restricts itself to just $b \ge 1$. $\endgroup$ May 8, 2019 at 7:26
  • $\begingroup$ @JohnOmielan It's just a natural restriction of the applied model from which the question arises. The model wouldn't really make sense for b<1. $\endgroup$ May 9, 2019 at 9:03

1 Answer 1

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You want to show that

$$((rt)^{1/b}-1)\log(r) - t^{1/b}(r^{1/b}-1)\log(rt) > 0 \tag{1}\label{eq1}$$

The LHS can be rewritten as

\begin{align} g(b) & = ((rt)^{1/b}-1)\log(r) - ((rt)^{1/b}-t^{1/b})(\log(r) + \log(t)) \\ & = (rt)^{1/b}\log(r) - \log(r) - (rt)^{1/b}\log(r) - (rt)^{1/b}\log(t) + t^{1/b}\log(r) + t^{1/b}\log(t) \\ & = -\log(r) - (rt)^{1/b}\log(t) + t^{1/b}\log(rt) \tag{2}\label{eq2} \end{align}

Differentiating wrt $b$ gives

\begin{align} g'(b) & = \frac{1}{b^2}(rt)^{1/b}\log(t)\log(rt) - \frac{1}{b^2}t^{1/b}\log(rt)\log(t) \\ & = \frac{\log(t)\log(rt)t^{1/b}}{b^2}\left(r^{1/b} - 1\right) \tag{3}\label{eq3} \end{align}

Since $0 \lt t \lt 1$, then $\log(t) \lt 0$. Also, $r \gt \frac{1}{t} \; \Rightarrow rt \gt 1$, so $\log(rt) \gt 0$. Next, as $t \lt 1$, then $\frac{1}{t} \gt 1$, so $r \gt 1$, giving that $r^{1/b} - 1 \gt 0$. As can be seen in \eqref{eq3}, for $b \ge 1$, there is the negative factor of $\log(t)$, but all other factors are positive, so $g'(b) \lt 0$. Since for any positive $c$, $\lim_{\; b \to \infty} c^{1/b} = 1$, \eqref{eq2} gives

$$\lim_{b \to \infty} g(b) = -\log(r) - \log(t) + \log(rt) = 0 \tag{4}\label{eq4}$$

As $g'(b)$ is always negative, but the limit of $g(b)$ is $0$, this means $g(b)$ is an always positive function which is decreasing towards $0$, i.e., $g(b) \gt 0$, thus proving \eqref{eq1}.

Update: The positive & negative conditions of the original derivative $f'(b)$ and the $g'(b)$ in \eqref{eq3} here hold not only for $b \ge 1$, but more generally for all positive $b$, so $f'(b) \gt 0$ is true for all $b \gt 0$.

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