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I am interested in proving that the following equation on the interval $x \in [c,1]$ is minimized either at the endpoints or where $x=\sqrt{c}$:

$f(x)=\frac{-1}{\log(1-x)}+\frac{-1}{\log(1-\frac{c}{x})}$

where $0<c<1$ is some constant.

It seems the first term of this equation is concave for $x > \frac{e^2-1}{e^2}$ and the second term of the equation is always concave so the sum is therefore also concave for $c > \frac{e^2-1}{e^2}$.

Where I am having trouble is when $c \leq \frac{e^2-1}{e^2}$ and $f(x)$ is now the sum of concave and convex functions. It seems intuitive that this should be minimized either at the endpoints or at the balance point when $x=\sqrt{c}$. I want to show that there are no other local minima other than $x = \sqrt{c}$, which is confirmed in all the plots I've made, but I'm not sure how to prove it analytically. (I've had trouble analyzing the derivatives directly because they get messy.)

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  • $\begingroup$ It would be sufficient to show that the fourth derivative of f is always negative (as that limits the number of inflections that can occur). This is true whenever c>0.45. But as you said, the algebra is messy. $\endgroup$
    – Alex Meiburg
    Commented Feb 13, 2019 at 21:21
  • $\begingroup$ @Alex Meiburg Do you know what happens for $c \le 0.45$? My not totally rigorous calculation, which could be subject to numerical difficulties, indicates that $f^{''''}(x) < 0$ for all $c > 0$ and $c \le x \le 1$. Specifically, I calculated $f^{''''}(x) $ using MAPLE, then maximized it over this range using the BARON "(almost) rigorous": global optimization solver, and the maximum came out negative. However, not all intermediate expressions were bounded, so it's possible this is not really the global maximum. $\endgroup$
    – Mark L. Stone
    Commented Jun 9, 2019 at 18:10

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