Let $\Delta{}ABC$ be an acute triangle and $O$ its orthocenter, so that $O$ is in the interior of $\Delta{}ABC$. Is it true that $OA+OB+OC$ is less than the sum of any two sides of $\Delta{}ABC$? In other words, do all of the following necessarily hold:
- $OA+OB+OC<AB+AC$
- $OA+OB+OC<AB+BC$
- $OA+OB+OC<AC+BC$?
This question grew out of my previous SE question that asks about a more general scenario where we only assume that $0<\measuredangle{}ACB<\frac{\pi}{2}$ and seek a point $P$ on the altitude from $C$ that satisfies the inequality $PA+PB+PC<AC+BC$. Restricting to acute triangles and the orthocenter, as this question does, seems interesting and specific enough to warrant its own question.
Best Answer
Credit to @VMF-er
Let $H$ be the orthocenter of the acute-angled $\triangle ABC$ ($H$ is traditionally more commonly used).
Let $E \in AB$ and $F \in AC$ such that $\angle{BHE}=\angle{CHF}=90°$.
Observe that $AEHF$ is a parallelogram:
$$AE+AF>HA$$
Further, $\triangle BEH$ and $\triangle CFH$ are right triangles:
$$BE>HB \;\;\text{and}\;\; CF>HC$$
Adding the three inequalities, we get the desired inequality:
$$HA+HB+HC<AB+AC$$
Applying this cyclically proves the problem. $\;\blacksquare$
In fact, the following is a corollary of this result (using the extended law of sines and $AH=2R\cos(A$)...:
$$\cos A+\cos B+\cos C<\sin B+\sin C$$
We could also reverse-engineer and use this trigonometric inequality as our starting point, but that would require using sum to product formulae and an analysis of increasing/decreasing behaviour on certain domains, etc. which seems like a clumsier solution.