Does the orthocenter of an acute triangle satisfy this inequality? 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.
 A: Hints:
Let the foots of altitudes on sides BC=a, AC=b and AB=c be D, F and E and altitudes be $h_a$, $h_b$ and $h_c$ respectively. Use this fact:
$h_a+h_b+h_c< a+ b+ c$
Or:
$(OA+ OD)+(OB+OE)+(OC+OF)<a+b+c$
For example move $a$ to LHS:
$OA+OB+OC+[(OE+OF+OD)-a]<a+b$
Now you have to show $(OE+OF+OD)-a\geq 0$
It is obvious for condition $a<b<c$ for biggest side c you may need some calculation to show that.
A: Using more traditional $H$ for the orthocenter and $O$ for the circumcenter.

Let $\rho,r$ and $R$ denote the semiperimeter, inradius and circumradius of the triangle, respectively
and WLOG let $c$ be the longest side of the triangle.
Hint: for acute triangles
\begin{align}
|AH|+|BH|+|CH|&=2(r+R)
\tag{1}\label{1}
.
\end{align}
Hence it's suffice to show that
\begin{align}
2(r+R)&<a+b
\tag{2}\label{2}
,\\
2(r+R)&<2\rho-c
\tag{3}\label{3}
,\\
c&<2\rho-2(r+R)
\tag{4}\label{4}
,\\
\frac cR&<2\frac\rho{R}-2\frac rR-2
\tag{5}\label{5}
.
\end{align}
For acute triangles
we have $\tfrac cR<2$
and
$\tfrac\rho{R}>\tfrac rR+2$,
that is
\begin{align}
2\frac\rho{R}-2\frac rR-2
&>2
\tag{6}\label{6}
,
\end{align}
hence \eqref{5} is true.
A: 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.
