Dividing a spherical cap into three equal wedges Background: Optimal ways to cut an orange.
In this problem, we have a spherical orange, and we do not wish to eat its central column which is modelled as a cylinder. Part of the procedure involves an initial cut down the orange tangential to the central column, which produces a spherical cap, then further splitting the cap into three wedges where each cut passes through the midpoint of the initial cut. The goal is to make these three wedges have equal volumes.
I was able to derive closed-form expressions for the volumes, but I noticed that it seemed impossible to have both equal volumes and equal angles ($\alpha_1=\alpha_2=\alpha_3=\pi/3$) of the three wedges at the same time. In my partial answer to the question, this scenario is equivalent to proving that the equation $$\small2\arctan s+\frac{3s}4(1-3s^2)-\frac{11+3s^2}8\sqrt{1-3s^2}\arctan\frac{2s}{\sqrt{1-3s^2}}=\frac\pi6(1-\sqrt{1-3s^2})^2(2+\sqrt{1-3s^2})$$ over $s\in(0,1/\sqrt3)$ has no roots. Note that equality holds when $s=0$ (we make no cuts at all which is an impractical solution), and when $s=1/\sqrt3$ (when there is no central column, and the cap is a hemisphere). I have been unable to prove this step due to the complexity of the equation but I suspect problems of this sort have already been investigated.
Is there an existing reference on this result - i.e. that $\alpha_1=\alpha_2=\alpha_3=\pi/3$ only when the spherical cap is a hemisphere?
 A: Let $f(s)$ stand for the difference between the left- and right-hand sides of your displayed inequality. We want to show that $f<0$ on the interval $(0,1/
\sqrt3)$.
Let
$$f_1(s):=f'(s)\frac{8 \sqrt{1-3 s^2}}{3 s \left(s^2+1\right)}
=9 \tan ^{-1}\left(\frac{2 s}{\sqrt{1-3 s^2}}\right)-\frac{6 s \left(-9 s^2+2 \pi  \sqrt{1-3 s^2}
   s+3\right)}{\sqrt{1-3 s^2} \left(s^2+1\right)}.$$
Then
$$f_1'(s)\frac{\left(s^2+1\right)^2}{24 s}=\frac{6 s}{\sqrt{1-3 s^2}}-\pi,$$
which is clearly $-+$ on $(0,1/\sqrt3)$ -- that is, $<0$ on $(0,c)$ and $>0$ on $(c,1/\sqrt3)$ for some $c\in(0,1/\sqrt3)$. So, $f_1$ is down-up on $(0,1/\sqrt3)$ -- that is, decreases on $(0,c)$ and increases on $(c,1/\sqrt3)$ for some $c\in(0,1/\sqrt3)$.
Also, $f_1(0)=0$ and $f_1(\frac1{\sqrt3}-)=3\pi/2>0$. So, $f_1$ is $-+$ on $(0,1/\sqrt3)$, and hence $f'$ is $-+$ on $(0,1/\sqrt3)$, which implies that
$f$ is down-up on $(0,1/\sqrt3)$.
Also, $f(0)=0$ and $f(\frac1{\sqrt3}-)=0$. Thus, $f<0$ on $(0,1/
\sqrt3)$, as desired. $\quad\Box$
