Given two points, $a, b$ inside a triangle $T$ whose perimeter is L we want to prove that the distance between them:
$$|d(a,b)| \leq L. $$
One could easily prove this graphically, ie. the distance two points in the triangle is obviously less than the total perimeter of the triangle, but how can I prove this formally?
On
Let $A$, $B$, and $C$ be the triangle's vertices. Let $D$ and $E$ be where the extensions of $Aa$ and $Ab$ intersect with $BC$. Then the triangle inequality gives: $$ d(a,b)\leq Ae + Ab\leq AD + AE\leq (AB+BD)+(AC+EC)\\ \leq AB + AC + (BD+DE+EC)=L. $$ (In the picture I have, labeling is such that on $BC$, the points are $B,D,E,C$ in that order from left to right. If your picture is different, adjust the last inequality in the first line above.)
In fact we have $$d(a,b)\leq \frac{L}{2}$$ Indeed, to see this, extend the line $(ab)$ so that it intersects the perimeter of the trinangle at two points $X$ and $Y$. Clearly $d(a,b)\leq d(X,Y)$. Now, we can go from $X$ to $Y$ moving on the perimeter of the triangle in two ways, and the shortest of these two paths is shorter than $L/2$ because the sum of their lengths is $L$. So, $d(X,Y)\leq L/2$ and we are done.