The length of a line segment within a triangle is less than the perimeter of the triangle. Intuitively the proposition is true, but how to prove it? By analytic geometry this boils down to proving the inequality
$\left[(\sum_{i=1}^3(t_i'-t_i)x_i)^2+(\sum_{i=1}^3(t_i'-t_i)y_i)^2\right]^{\frac{1}{2}}\leq\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}+\sqrt{(x_1-x_3)^2+(y_1-y_3)^2}+\sqrt{(x_2-x_3)^2+(y_2-y_3)^2}$
where $\sum_{i=1}^3t_i'=\sum_{i=1}^3t_i=1$ and $t_i',t_i\geq 0,i=1,2,3,$
and the inequality is strict when the triangle in question is a proper one. But I can't prove it.
 A: If $a = t_1 x + t_2 y + t_3z$ and $b = s_1 x + s_2y + s_3z$ are two points in the triangle then
$$
 \Vert a - b \Vert = \Vert (t_1 - s_1) x + (t_2 - s_2) y + (t_3 - s_3) z \Vert
 $$
Since $t_3 - s_3 = (1-t_1-t_2) - (1- s_1-s_2) = -t_1-t_2+s_1+s_2$, this is equal to
$$
\begin{align}
 \Vert a - b \Vert &= \Vert (t_1 - s_1) (x-z) + (t_2 - s_2) (y-z)  \Vert \\
 &\le |t_1 - s_1| \cdot \Vert x - z \Vert + |t_2 - s_2| \cdot \Vert y - z\Vert \\
  &\le  \Vert x -z \Vert + \Vert y -z \Vert \\
  &\le \Vert x -z \Vert + \Vert y -z \Vert + \Vert x -y\Vert \, ,
\end{align}
$$
using the triangle(!) inequality for the Euclidean norm, and that $-1 \le t_i - s_i  \le 1$ for $i =1, 2, 3$.
The inequality is strict unless $x=y=z$.

A better estimate is possible: At least two of the three numbers $t_i - s_i$ must have the same sign. Without loss of generality assume that $t_1-s_1$ and $t_2 - s_2$ have the same sign. Also let $$
 L = \max(\Vert x-z \Vert, \Vert y-z \Vert, \Vert x-y \Vert)
$$
be the longest side length. Then the above calculation shows that
$$
\begin{align}
\Vert a-b \Vert &\le |t_1 - s_1| \cdot \Vert x - z \Vert + |t_2 - s_2| \cdot \Vert y - z\Vert \\
&\le \left( |t_1 - s_1| + |t_2 - s_2|\right) \cdot L \\
&= \left( |(t_1 - s_1) + (t_2 - s_2)|\right) \cdot L \\
&= \left( |s_3 -t_3|\right) \cdot L \\
&\le L \, ,
\end{align}
$$
i.e. the length of every segment is at most the length of the longest triangle side. It follows that
$$
\Vert a-b \Vert \le \frac 12 \bigl( \Vert x -z \Vert + \Vert y -z \Vert + \Vert x -y\Vert \bigr) \, .
$$
