Apologies if I´m not using the correct mathematical wording or notation. I´ll try my best in the following problem

Suppose you have a coordinate system and 5 coordinates ($A1$, $A2$, $B1$, $B2$, $M_{x,y}$) whereas $M_{x,y}$ is variable and the others are fixed. The distances between them are measured in Euclidean distances.

Person A wants to go from $A1$ to $A2$

Person B wants to go from $B1$ to $B2$

But Person A needs to meet Person B at a variable meeting point $M_{x,y}$

Let $D(C1,C2)$ be the Euclidean distance between two coordinates $C1$ and $C2$.

Let $W(M_{x,y},A1,B1)$ be the waiting time of Person A for Person B, or Person B for Person A (whoever arrives first/has the shorter distance) at meeting point $M$, depending on the location of $A1$ and $B1$

Then we have the following distances

$$D(A1,M_{x,y})= \sqrt{(M_x-A1_x)^2+(M_{y}-A1_y)^2}$$ $$D(M_{x,y}, A2)= \sqrt{(A2_x-M_x)^2+(A2_y-M_y)^2}$$

$$D(B1,M_{x,y})= \sqrt{(M_x-B1_x)^2+(M_y-B1_y)^2}$$ $$D(M_{x,y}, B2)= \sqrt{(B2_x-M_x)^2+(B2_y-M_y)^2}$$

and the following waiting time $$W(M_{x,y},A1,B1) = \max(D(A1,M_{x,y})-D(B1,M_{x,y}),D(B1,M_{x,y})-D(A1,M_{x,y}))$$

Then the the objective function would be $$Z = D(A1,M_{x,y}) + D(M_{x,y},A2) + D(B1,M_{x,y}) + D(M_{x,y},B2) + W(M,A1,B1)$$

I have plotted this for varying coordinates for $M_{x,y}$ and see that it is a convex function. I´d like to prove that. Here is how I tried to do it:

I want to build the second derivative and show it is greater 0. As I have two variables (x and y coordinate) I´d need to use Hessian Matrix and partial derivatives. But the function $W(M_{x,y}$ is a max() function, which is somewhat problematic (at least to me) for derivations. My idea is that in fact it never makes sense to wait as it would always be better to move a little closer to the other Person. As a result I determine the equation that separates A1 and B1 such that the distance from A1 and B1 to the equation is always the same. Now I know that M(x,y) must be on this equation $M(y) = mx + b$. I derive the equation in the following way:

Middle point between $A1$ and $B1$

$$M_\text{Middle}= \left(\frac{A1_x+B1_x}2, \frac{A1_y+B1_y}2\right)$$

Slope $m$ of equation $M(y)$

$$m=-\frac{B1_x-A1_x}{B1_y-A1_y}$$

Using $M_\text{Middle}$ and $m$ in $M(y)$ I get for $b$

$$b = \frac{A1_y+B1_y}2+\frac{B1_x-A1_x}{B1_y-A1_y}\frac{A1_x+B1_x}2 = \frac{B1_y^2-A1_y^2+B1_x^2-A1_x^2}{2(B1_y-A1_y)}$$

The final equation should be:

$$M(y) = -\frac{B1_x-A1_x}{B1_y-A1_y}x+ \frac{B1_y^2-A1_y^2+B1_x^2-A1_x^2}{2(B1_y-A1_y)}$$

Not I can insert this equation into $Z$ and would get

\begin{align} Z = {} & \sqrt{(M_x-A1_x)^2+ \left(-\frac{B1_x-A1_x}{B1_y-A1_y}x+ \frac{B1_y^2-A1_y^2+B1_x^2-A1_x^2}{2(B1_y-A1_y)}-A1_y\right)^2} \\[6pt] & {} + \sqrt{(A2_x-M_x)^2+\left(A2_y--\frac{B1_x-A1_x}{B1_y-A1_y}x+ \frac{B1_y^2-A1_y^2+B1_x^2-A1_x^2}{2(B1_y-A1_y)}\right)^2} \\[6pt] & {} + \sqrt{(M_x-B1_x)^2+\left(-\frac{B1_x-A1_x}{B1_y-A1_y}x+ \frac{B1_y^2-A1_y^2+B1_x^2-A1_x^2}{2(B1_y-A1_y)}-B1_y\right)^2} \\[6pt] & {} + \sqrt{(B2_x-M_x)^2+\left(B2_y--\frac{B1_x-A1_x}{B1_y-A1_y}x+ \frac{B1_y^2-A1_y^2+B1_x^2-A1_x^2}{2(B1_y-A1_y)}\right)^2} \end{align}

Now I "only" need to build the first and second derivative based on $M_{x}$. I could use the chain rule, which gets ugly but I manage to do it. However, the derivative is so long/complicated, that I could not tell if this is greater 0 or not.

Do you have any idea how the problem could be simplified?