Motivation behind Arithmetic Mean I know that the arithmetic mean $(x_1+x_2+...+x_n)/n$ is the value that minimizes $f(x)=\sum_{k=1}^n (x_k-x)^2$; however, I'm looking for an intuitive relationship between the mean and $g(x)=\sum_{k=1}^n \left\vert{x_k-x}\right\vert$.
I'm aware that the mean doesn't necessarily minimize $g$; instead, one of the ${x_k}'s$ does (although maybe not uniquely), but I noticed that, using the triangle inequality, $$h(x)=n\left\vert{x-((x_1+x_2+...+x_n)/n)}\right\vert\le\sum_{k=1}^n\left\vert{x_k-x}\right\vert$$
and the value that minimizes $h$ is the arithmetic mean. That is, the point at which the minimum of this lower bound function of $g$ occurs is the mean. However, I'm unable to think of an intuitive explanation for this. Perhaps it's just a coincidence? Basically, I'm looking for any intuitive connections between the mean and the sum of absolute differences. Any ideas/comments would be greatly appreciated!

At it's most basic and intuitive: The arithmetic mean isn't about the absolute difference, it's about the actual difference:
$\sum_{k=1}^n ({x_k-\bar{x}}) = (x_1 - \bar{x}) + (x_2 - \bar{x}) + (x_3 - \bar{x}) + ... + (x_n - \bar{x})$
$= x_1 + x_2 + x_3 + ... + x_n - n*\bar{x}$
If $\sum_{k=1}^n ({x_k-\bar{x}}) = 0$, then $x_1 + x_2 + x_3 + ... + x_n = n*\bar{x}$
$\bar{x} = \frac{x_1 + x_2 + x_3 + ... + x_n}{n}$
Going by the absolute differences instead will not be this simple.
Have you gotten into Standard Deviations yet? The Standard Deviation is a measure of how different each piece of data is from the mean, and even that is based on $\sqrt{\sum_{k=1}^n ({x_k-\bar{x}})^2}$ instead of on $\sum_{k=1}^n \sqrt{({x_k-\bar{x}})^2}$
