Median of the sum vs. sum of the median for Gaussian variables This is a problem I stumbled upon in my research. Consider $n$ Gaussian random variables $x_i \sim \mathcal{N} (\mu_i, \sigma_i^2)$, each with its own mean $\mu_i$ and variance $\sigma_i^2$. Can we say that
\begin{equation}
\text{Median} \left( \sum_{i=1}^n x_i^2 \right) \ge \sum_{i=1}^n \text{Median} \left( x_i^2 \right) \; ?
\end{equation}
If the $x_i$ were standard normal variables ($\mu_i = 0$ and $\sigma_i^2=1$) we could compute $\text{Median} (x_i^2) \simeq 0.4549$, and the sum $\sum_{i=1}^n x_i^2 $ would follow the $\chi^2$-distribution with $n$ degrees of freedom, which is reported to have median $\simeq n \left( 1 - \frac{2}{9n}\right)^3$. The above inequality would be
\begin{equation}
\left( 1 - \frac{2}{9n}\right)^3 \ge 0.4549
\end{equation}
which is satisfied $\forall n \in \mathbb{N}$. The question is whether the inequality is true even for non standard normal variables. The fact that it holds for standard normal variables is related to the heavy-tailedness of the distribution of $x_i^2$, so I think it might hold also for general Gaussian variables.
 A: Here is a simple proof, by comparison with appropriate symmetric variables.
For a normal variable $X$, let $m$ be the median of $X^2$. The graphs show $X\sim N(4,1)$.
Now we take a variable $Y$ which is a version of $X^2$ but symmetrized from right to left about $m$:


In formulas:
$$
f_Y(y) =
\begin{cases}
f_{X^2}(y)\phantom{2m-\, }\ \text{ if }\, y>m\\
f_{X^2}(2m-y)\ \text{ if }\, y<m\\  
\end{cases}$$
$$F_Y(y) =
\begin{cases}
\phantom{1-\, }F_{X^2}(y)\phantom{2m-\, }\ \text{ if }\, y>m\\
1-F_{X^2}(2m-y)\ \text{ if }\, y<m\\  
\end{cases}
$$
Since the $Y_i$'s are symmetric with median $m_i$, the values near $(y_1, \ldots y_n)$ and $(2m_1-y_1, \ldots 2m_n-y_n)$ are equally probable and on opposite sides of $m_1+\cdots+m_n$. So $m_1+\cdots+m_n$ must be the median of their sum.
Also $X^2$ dominates $Y$, i.e. $F_{X^2}(y)\le F_Y(y)$, as can be seen in graphs like the above. Thus:
$$
\begin{align}
\text{median}\left(\sum X_i^2\right)
&\ge \text{median}\left(\sum Y_i\right)\ &\text{ (by the dominance of the }X\text{'s)}\phantom{\ \square}\\
&=\sum\text{median}(Y_i)\ &\text{ (by the symmetry of the }Y\text{'s)}\phantom{\ \square}\\
&=\sum\text{median}(X_i^2)\ &\text{ (by the construction of the }Y\text{'s)}\ \square
\end{align}
$$
