Variation of the sum of distances

Question

Let $l$ be a line and $A$ and $B$ two points on the same side of $l$. To find the point $P$ for which $AP+PB$ is minimum we take the intersection of $l$ and the line joining $B$ and the symmetric $A'$ of $A$ with respect to $l$. For any point $M$ of $l$ other than $P$ we have $AM+MB=A'M+MB>A'B=AP+PB$ so $P$ is the desired point.

My question is $\color{red}{\text{how to prove that $AM+MB$ increases with $PM$?}}$.

My attempt: If $M'$ is another point of $l$ such that $PM'>PM$ then $AM<AM'+MM'$ and $BM<BM'+MM'$. I want to prove that $AM+MB<AM'+M'B$ using only the triangle inequality.

Answer 1

Sometimes a figure is worth a thousand words:

This distance from $A$ to $B$ via the line is the same as the distance from $A$ to $B'$... the shortest of which is a straight line. Using the elementary fact from Euclidean geometry that the shortest distance between two points is a straight line we see: $\color{red}{\text{all other such paths (see green) must be longer.}}$ Proof completed.

Answer 2

Your inequality is false: see diagram below for a counterexample.

Answer 3

Let $M$ and $M'$ be two point on line $l$ on the same side of $P$ such that $PM<PM'$ and let $C$ be the intersection point of $BM$ and $A'M'$. By the triangle inequality: $$MA'<MC+CA' .............(1)$$ and $$CB<CM'+M'B .............(2)$$ Now add $MB$ to both sides of $(1)$ and $CA'$ to both sides of $(2)$ to obtain $$\color{red}{MA+MB}=MA'+MB<CA'+CB\color{red}{<}M'A'+M'B=\color{red}{M'A+M'B}$$