Is the variance of the sample median of iid variables at most the variance of each variable?

At least for a symmetric parent $f$ the variance of the sample median is at most $\sigma^2$. Consider integrable densities with finite moments of order two. By normalization (i.e., by going from ${\bf X}$ to ${\bf X}' = ({\bf X} - \mu)/\sigma$) we may assume without loss of generality that $\mu=0$ and $\sigma^2=1$. For a symmetrical parent, $f(x)=f(-x)$, we have by standard partial integration
\begin{eqnarray*} \mbox{E}[{\bf X}^2] & = & -2 \int_{-\infty}^{0} \, x F(x) \, dx \; + \; 2 \int_{0}^{\infty} \, x [1-F(x)] \, dx \end{eqnarray*} which under the assumption of symmetry and standardization simplifies to \begin{eqnarray*} \sigma^2 & = & 4 \, \int_{0}^{\infty} \, x [1-F(x)] \, dx \; = \; 1 \end{eqnarray*} Looking at samples of size $n=3$, the density $g$ of the second order statistic, that is, of the sample median, equals $g(x) = 6 f(x) F(x) [1-F(x)]$. It is therefore symmetric, too, and the distribution function is $G(x) = 3 F^2(x) -2F^3(x)$. Note that by assumption $F(0)=\frac{1}{2}$. For $x \geq 0$ (i.e., for $\frac{1}{2} \leq F \leq 1$) it is elementary to show that \begin{eqnarray*} 1 - G(t) \; = \; 1 - 3 F^2(x) + 2F^3(x) & \leq & 1-F(x) \end{eqnarray*} Thus, we get for the variance of the sample median \begin{eqnarray*} 4 \, \int_{0}^{\infty} \, x [1-G(x)] \, dx & \leq & 4 \, \int_{0}^{\infty} \, x [1-F(x)] \, dx \; = \; \sigma^2 \end{eqnarray*} The variance of the sample median cannot increase with $n$, and in any case an essentially similar proof may be given for any odd sample size $2m-1$ using the density of the $m-$th order statistic. The bound $\sigma^2$ for the variance of the sample median may be attained arbitrarily close, for example, by a parent $f$ that is an even mixture of two unit-variance normals with widely separated means $\pm \mu$.