Let $0<u<\sin x<x<\frac{\pi}{2}$, and$$f(x)=\frac{x}{u}\left(\frac{1}{(\sin x) - u}-\frac{1}{(\sin x) + u}\right)-\frac{1}{x-u}-\frac{1}{x+u}.$$Show that $f(x)$ as a function of $x$ with fixed $u$ is convex.
We can also write$$f(x)=\frac{2x}{(\sin^2x)-u^2}-\frac{2x}{x^2-u^2}.$$
I have calculated$$f''(x) = \frac{2x\cos^2x}{u}\left(\frac{1}{((\sin x) -u)^3}-\frac{1}{((\sin x) + u)^3}\right) + \frac{x(\sin x)-2\cos x}{u}\left(\frac{1}{((\sin x) -u)^2}-\frac{1}{((\sin x) + u)^2}\right) -\frac{2}{(x-u)^3}-\frac{2}{(x+u)^3}.$$
Plots of the two-variable function support the conclusion.
I have been able to show$$\lim_{u\to 0^+}f''(x;u) = \frac{12x\cos^2x}{\sin^4x} - \frac{8\cos x}{\sin^3x} + \frac{4x}{\sin^2x} - \frac{4}{x^3}>0.$$