Prove or disprove an inequality involving statistics Do we have any result in statistics like this:

$$|\overline x - \mu_e| \leq \sigma$$

Here $\overline x$ denotes the usual mean of some given discrete observations, $\mu_e$ their median and $\sigma$ the standard deviation of the variables. I got this feeling because I found the following problem:

There are $2n+1$ numbers such that $x_1 \leq x_2 \leq x_3 \cdots \leq x_{2n} \leq x_{2n+1}$ and $\sum_{i=1}^{2n+1} x_i =0$. Show that, $$ x_{n+1}^{2} \leq \frac {1}{2n+1} \sum_{k=1}^{2n+1} x_{k}^{2}$$

plus, $\overline x$ and $\mu_e$ are supposed to be measures of central tendencies and $\sigma$ is supposed to find out how variables are scattered from the mean. It would be really strange if two such values be more far than the average scattering of all variables. 
Though here are two questions, my priority is the first one. But that does not mean I would not appreciate any other methods of solving the second one. Thank you in advance!
 A: There are many proofs of your first statement, which is true for any random variable $X$ with mean $\mu$, median $m$ and finite variance $\sigma^2$, discrete or continuous.  Here is one:


*

*Suppose $\mu \ge m$ (otherwise consider $Y=-X$).  

*Let $p=\Pr(X \le m)$: since $m$ is the median we have $p \ge \frac12 \ge 1-p = \Pr(X \gt m)$. 

*Let $L=E[X\mid X \le m]$ and $H=E[X\mid X \gt m]$ so $\mu=pL+(1-p)H$. We have $L\le m$ and $H \ge 2\mu-m$.

*$E[(X-\mu)^2\mid X \le m] = Var(X\mid X \le m) + (\mu-L)^2 \ge (\mu-m)^2$ and similarly  $E[(X-\mu)^2\mid X \gt m] = Var(X\mid X \gt m) + (H-\mu)^2 \ge (\mu-m)^2$.

*So $\sigma^2 = Var(X)=E[(X-\mu)^2] \ge p(\mu-m)^2+(1-p)(\mu-m)^2= (\mu-m)^2$ and thus taking square roots $ |\mu-m| \le \sigma$.


If you have a sample, this statement would be true if it were treated as a population.  It is still true if you were to use the $\frac1{n-1}$ definition of sample standard deviation, since that is $\frac{n}{n-1}$ times, i.e. greater than, the population definition of standard deviation. 
Your second expression is a corollary of your first, since $x_{n+1}$ is the median while the mean is zero and the variance is the right-hand side, so it is simply the square of $ |\mu-m| \le \sigma$. 
