Consider the following simplex:
$$\Delta_N = \left\{x \in \mathbb{R}^N : x_i \geq 0 \wedge \sum_{i=1}^N x_i= 1\right\}.$$
Find the maximum euclidean distance that can be obtained by choosing $2$ different vectors $x$ and $y$ in $\Delta_N.$
The intuition suggests me that one of these $2$ vector should correspond to one vertex of the simplex. For example, $x =[1, 0, \ldots, 0].$ Moreover, I would say that the farthest vector from $x$ is $y = [0, \frac{1}{N-1}, \ldots, \frac{1}{N-1}]$, which is the middle point of the face of the simplex opposite to $x$. Just figure out a regular polyhedron to arrive to such intuition.
In this case, the distance is:
$$\|x - y\|=\sqrt{(1-0)^2 + \sum_{i=2}^N \left(0 - \frac{1}{N-1}\right)^2} = \sqrt{\frac{N}{N-1}}.$$
As I said before, this is just an intuition! I've tried to obtain a solution by looking for the maximum of the function $\|x - y\|$ using derivatives, but I had no success. For example, I can fix $x = [1, 0, \ldots, 0]$ and let $y$ vary. Using the relation $y_1 = 1 - \sum_{i=2}^Ny_i$, the gradient with respect to $y_2, y_3, \ldots, y_N$ is a $(N-1)$-vector composed by non-negative components. The gradient is $0$ only for $y = x$, which obviously is the minimum distance possible.
Do you have any idea about how to approach this problem?
Well, you know that $$\Vert x-y \Vert ^{2} = \Vert x \Vert ^{2} +\Vert y \Vert ^{2}- 2\langle x,y\rangle,$$ So, a trivial upper bound here for squared distance is 2, as $\Vert x \Vert, \Vert y \Vert \leq 1$, and $\langle x,y\rangle \geq 0$.
Do any two points achieve this bound? Sure, choose x = $e_i$ and $y=e_j$, (the ith and jth standard basis vectors) for $i \neq j$.