I'm talking specifically of population SD, where $$s = \sqrt{ \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2}$$
I have a hunch that $s$ is maximized for a certain mean $\bar{x}$ when the values in our set are symmetric at the extremes of the range $[a,b]$ such that $\bar{x} = \frac{a+b}{2}$.
For example {1, 1, 1, 100, 100, 100}. (not sure about odd number of values, though).
The idea is that I ultimately want to prove that the Standard deviation cannot be greater than the greatest distance from any value to the mean.
I feel if I can show how to max $s$ and prove it, it'll make the larger proof much easier to deal with algebraically, since all I'll have to do is show that when the values are symmetric in that way, $s$ is equal to the greatest distance. And since that's the max that $s$ can be, it should prove that it can never be greater than the greatest distance in our set.
Obviously if $d$ is the greatest distance from any of the values to the mean, then every $(x_i - \bar{x})^2 \le d^2$; plug into the formula for SD to find that SD $\le d$. Moreover, it can only be $d$ when all distances are equal to $d$, no less; and since the average must still be $d$, this means half the values are at one extreme and half at the other. And this max can't be attained for an odd number of values.
However, this assumes that the interval given ($[a,b]$) and the mean $\bar{x}$ are not independent; rather, $\bar{x}$ is the midpoint of the range. What if you are looking for values in $[0,1]$ but want the maximum SD for average 0.4, not 0.5?