Given real numbers $x_1, x_2, \ldots x_n$, the mean $\overline{x} = \frac{x_1, x_2, \ldots x_n}{n}$ is the unique value which minimizes $\sum (x_i - \overline{x})^2$.
I know how to prove this with calculus (take the derivative of the sum with respect to $\overline{x}$ and set to zero, the mean pops out) but I'm looking for a more conceptual understanding of this fact, and I think the calculus is getting in the way of my conceptual understanding.
Can we prove it without calculus? Ideally, without too much algebra — perhaps geometrically?
I hope this does not count as "too much" algebra but note that for any real $y$, we have \begin{align} \sum(x_i - y)^2 &= \sum(x_i^2 - 2x_iy + y^2)\\ &= ny^2 - 2\left(\sum x_i\right)y + \left(\sum x_i^2\right)\\ &= ny^2 - 2Ay + B\\ &= n\left(y - \dfrac{A}{n}\right)^2 + \left(B - \dfrac{A^2}{n}\right) \end{align}
Note that $A, B, n$ are independent of $y$. Thus, the RHS is clearly minimised when the term $y - A/n$ is zero. (The minimum value of a perfect square.)
This gives you that $y = A/n$. Noting that $A = \sum x_i$, we see that $y = \bar{x}$.
Additional remark: Not really a geometric proof but the result can be interpreted as saying that the moment of inertia is minimum about axes passing through the centre of mass. (Which can be seen to be true as a result of the parallel axis theorem but I fear that that is too much physics for my own liking...)