I'm thinking there are $n$ numbers, call them $a_1$ through $a_n$ … with mean $x$ and median $y$. I think it is true that $$\displaystyle\sum\limits_{k=1}^{n} (x-a_k)^2 \leq \displaystyle\sum\limits_{k=1}^{n} (y-a_k)^2$$
In words, this means the sum of squared differences from the mean is never bigger than the sum of squared differences from the median.
I have tested many sets of numbers and this is true for everything I tested. However, I do not know how to prove it.
Best Answer
Hint: Consider $ f(x) = \sum (x - a_k)^2 = nx^2 - (2 \sum a_k)x + \sum a_k^2 $.
When is the minimum of $f(x)$ achieved?