Let us consider $4$ Points $A,B,C,D$ on a line $g$.
We are to prove: If $\vert AP\vert+\vert PD\vert >\vert BP\vert+\vert CP\vert $ for all $P$ not on $g$, then $\vert AB\vert=\vert CD\vert$.
I solved the question using the contraposition of the statement and calculus: We will first show that the inequality can be reversed for a point $P$ on $g$. Let us consider a point further to the right on $g$ then $D$, and let it be $P$; additionally, let $z>x$. We define $\vert AB\vert=x, \; \vert BC\vert=y, \;\vert CD\vert=y, \; \vert DT \vert=w$. Then, if the inequality were still to hold, we must have: $\vert AP\vert+\vert PD\vert >\vert BP\vert+\vert CP\vert$ or, $x+y+z+2t\geq 2z+y+2t$ or $x>z$, a contradiction.
We now choose $P$ such that for its foot $T$ on $g$, we have $\vert DT\vert=w$. Let $\vert TE\vert=\epsilon$. Now, using the pythagorean theorem, we must have: $\vert AP\vert+\vert PD\vert =\sqrt{(x+y+z+w)^2+\epsilon^2}+\sqrt{z^2+\epsilon^2}>\sqrt{(y+z+w)^2+\epsilon^2}+\sqrt{(z+w)^2+\epsilon^2}$ Letting $\epsilon$ approach $0$, by continuity we find a point of equality and thus for the inequality to hold, we must have $x=z$.
While this approach works it is definetly not the most elegant, because the problem was posed in a competition for middle schoolers. What is an elementary solution?
WLOG assume $x<z$. Place P on top of D such that $PD \perp AD$.
Let $\angle PAB = \alpha, \angle PBC=\beta, \angle PCD = \gamma$. Then $0 <\alpha < \beta < \gamma <\frac{\pi}{2}\Rightarrow \cos \alpha > \cos \beta > \cos \gamma.$
Now, $$ |AP| = \frac{x+y+z}{\cos \alpha}, |BP|=\frac{y+z}{\cos \beta},|CP|=\frac{z}{\cos \alpha}, |DP| = z \tan \alpha. $$
$$ 0 < |AP|+|DP|-|BP|-|CP|=(|AP|-|BP|) - (|CP|-|DP|)\\ = \left(\frac{x+y+z}{\cos \alpha} - \frac{y+z}{\cos \beta}\right)-\left(\frac{z}{\cos \gamma} - \frac{z\sin \gamma}{\cos \gamma}\right)\\ < \left(\frac{x+y+z}{\cos \beta} - \frac{y+z}{\cos \beta}\right)-\frac{z(1-\sin \gamma)}{\cos \gamma}\\ = \frac{x}{\cos \beta} -\frac{z(1-\sin \gamma)}{\cos \gamma} <\frac{x}{\cos \gamma} -\frac{z(1-\sin \gamma)}{\cos \gamma} $$
If we choose $\gamma$ very close to 0, such that $x < z(1-\sin \gamma)$ then $0 < \frac{x}{\cos \gamma} -\frac{z(1-\sin \gamma)}{\cos \gamma} < 0, \rightarrow \leftarrow$
Therefore $x=z.\blacksquare$