# Show $f'(b)>0$ for all $b\geq1$

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 $$0, and $$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 $$0, $$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 $$0, which I know is crucial, as it's easy to show this expression is less than zero for several cases where $$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.

• 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 $b \ge 1$. May 8, 2019 at 7:26
• @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. May 9, 2019 at 9:03

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 $$b \gt 0$$.