The following seems kind of obvious:
Let $\Delta ABC$ be a triangle in the plane and $P, Q$ two points in $\Delta ABC$ (in the interior or on the boundary). Then $|PQ| \le l$ where $l$ is the length of the longest triangle side.
Here $|PQ|$ denotes the Euclidean distance from $P$ to $Q$.
I can prove it (see below), but that proof does not provide much insight why this estimate is true. Is there a (simpler?, perhaps geometric?) proof which makes the inequality really obvious?
The fact that the triangle is convex probably plays a role, but convexity alone is not sufficient: The above is wrong for convex polygons with more than three sides.
My proof: Using barycentric coordinates we have
$$ P = p_1 A + p_2 B + p_3 C \\ Q = q_1 A + q_2 B + q_3 C $$ where the $p_i, q_i$ are non-negative real numbers with $\sum_{i=1}^3 p_i = \sum_{i=1}^3 q_i = 1$. At least two of the three differences $p_i - q_i$ must have the same sign, without loss of generality we can assume that $p_1-q_1$ and $p_2 - q_2$ have the same sign (both are non-negative or both are non-positive).
Using $\Vert \cdot \Vert$ for the Euklidean norm now, we have $$ \begin{align} \Vert P-Q \Vert &= \Vert (p_1-q_1) A + (p_2-q_2)B + (p_3-q_3)C \Vert \\ &= \Vert (p_1-q_1) (A-C) + (p_2-q_2)(B-C)\Vert \\ &\le |p_1 - q_1| \cdot \Vert A-C \Vert + |p_2 - q_2| \cdot \Vert B-C \Vert \\ &\le \bigl( |p_1 - q_1| + |p_2 - q_2| \bigr) \cdot l \\ &= |(p_1 - q_1) + (p_2 - q_2)| \cdot l \\ &= |p_3 - q_3 | \cdot l \\ &\le l \, . \end{align} $$
It's enough to solve our problem for $P\equiv A$ and $Q\in BC$.
If $b$ or $c$ is a greatest side, it's obvious.
Let $a$ be a greatest side.
Thus, there are $x\geq0$, $y\geq0$ such that $x+y=1$ for which: $$PQ=|\vec{PQ}|=|x\vec{AB}+y\vec{AC}|\leq xc+yb\leq xa+ya=a.$$