I'm looking to show $|{\rm med}(x)-\bar{x}|\le{\rm sd}(x)$. I did a bunch of simulations and the statement seems right to me.
$$ {\rm Var}(x)=\frac{1}{n}\sum\left(x_i-\bar{x}\right)^2=\frac{1}{n}\sum\left((x_i-{\rm med}(x))+({\rm med}(x)-\bar{x})\right)^2\\ =\frac{1}{n}\sum(x_i-{\rm med}(x))^2+\frac{1}{n}2\sum(x_i-{\rm med}(x))({\rm med}(x)-\bar{x})+({\rm med}(x)-\bar{x})^2. $$
So I hope to show
$$ \frac{1}{n}\sum(x_i-{\rm med}(x))^2+\frac{1}{n}2\sum(x_i-{\rm med}(x))({\rm med}(x)-\bar{x})\le0. $$
That is $$ \frac{1}{n}\sum(x_i-{\rm med}(x))(x_i-\bar{x}+{\rm med}(x)-\bar{x})\le0. $$
But it does not seem right because $(x_i-{\rm med}(x))(x_i-\bar{x})$ is only negative when $x_i$ is between ${\rm med}(x)$ and $\bar{x}$.
Is $|{\rm med}(x)-\bar{x}|\le{\rm sd}(x)$ even correct?
