For points $O,A,B,C$ in $\mathbb{R}^{3}$, I was trying to show $\angle AOC \le \angle AOB +\angle BOC$.
I could show this when all angles were acute.
First, I set $O$ to be the origin and $A,C$ to be on the first quadrant of $XY$ plane and let $B'$ to be the projection of $B$ on the $AOC$ plane and then dilated things away from $O$ to satisfy $|OA|=|OB|=|OC|=|OB'|=1$. Then making use of the fact that $OB'$ doesn't have third component, I verified using dot product formular for coordinates that $OA \cdot OB\le OA\cdot OB'$ and $ OC\cdot OB\le OC\cdot OB'$, hence obtaining $\angle AOB' \le \angle AOB$ and $\angle B'OC \le \angle BOC$ to yield $\angle AOC=\angle AOB' +\angle B'OC\le\angle AOB +\angle BOC$.
But I couldn't apply this approach to other cases, when angles could be obtuse or $B'$ could lie on $\overline{OA}$ or $\overline{OC}$.
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1 Answers
To work with the dot product, you will need to keep track of how the value of $\cos(x)$ varies as the angle $x$ increases.
Here's another approach. Take $O$ to be the origin. Rotate until $A$ lies on the positive $x$-axis. By reflecting if need be, further ensure that $B$ lies in the upper half space.
Now we break into cases. If $C$ lies in the upper half space, either $AOC\leq AOB$, in which case we are done, or $AOC>AOB$, in which case $AOC=AOB+BOC$ and we are done.
If instead $C$ lies in the lower half space. Take $D$ to be a point in the lower half space on the line through $O$ and $B$. If, measuring clockwise, $AOC\leq AOD$ then $BOC=AOB+AOC$, and we are done. Finally, if, measuring clockwise, $AOC> AOD$ then $AOB+BOC>180^{\circ}\geq AOC$, and again we are done.
A different approach would be to use complex numbers. With $O$ at $0$, $$AOB=\min\{|\theta|:\exists r>0, B=re^{i\theta}A\}.$$ Using this you can turn the question into one about the triangle inequality for real numbers.
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