# Minimising the total distance moved by two points in $\mathbb{R}^2$

Consider the set $S$ whose elements are pairs of smooth paths $A,B: [0,1]\rightarrow\mathbb{R}^2$ satisfying:

i) $A(0)= (0,0)$ and $B(0)=(1,0)$, and

ii) $B(1) - A(1) = (0,1)\in\mathbb{R}^2$.

What is the infimum of the set of distances

$$\left\{\int_A |A'(t)|\,dt + \int_B |B'(t)|\,dt:(A,B)\in S\right\}?$$

In other words, what is the smallest total distance that two points in the plane that are separated by distance $1$ must move through in order to become vertically separated and by the same distance?

My guess is that the answer is $\sqrt{2}$, but I'm not sure how to prove it.

The path from $A_0$ to $A_1$ has length at least $A_0 A_1$, same for the other one. So you have to show that $A_0 A_1 + B_0 B_1 \ge \sqrt{2}$.
Let $C=(1,-1)$. We have $B_0 B_1 = CA_1$. So $A_0 A_1 + B_0 B_1 = A_0 A_1 + C A_1 \ge A_0 C = \sqrt{2}$.
By defining $A_0(t) = A(t) - A(t) = (0,0)$ and $B_0(t) = B(t) - A(t)$, the requirements translate to $B_0(0) = (1, 0)$ and $B_0(1) = (0,1)$. We happen to have $(A_0, B_0)\in S$, and in this special case, it's clear that $\sqrt 2$ is the smallest value we can get. We thus have: $$\sqrt 2 \leq \int_0^1 \left(|A_0'(t)| + |B_0'(t)|\right)dt \\ = \int_0^1\left(0 + |B'(t) - A'(t)|\right)dt\\ = \int_0^1|B'(t) - A'(t)|\,dt\\ \leq \int_0^1 \left(|A'(t)| + |B'(t)|\right)dt$$