A Weighted Gaussian Inequality: $E[\frac{\sigma_n^2 x_n^2}{\sum_{i=1}^n \sigma_i^2x_i^2} ] \ge \frac{\sigma_n^2}{\sum_{i=1}^n \sigma_i^2}$ Given $\sigma_1 \ge \dots \ge \sigma_n \ge 0$,
and independent random gaussian variables $x_1, \dots, x_n \sim \mathcal N(0,1)$,
I want to show:
$$
\mathbb E\left[
\frac{\sigma_n^2 x_n^2}{\sum_{i=1}^n \sigma_i^2 x_i^2}
\right]
\ge \frac{\sigma_n^2}{\sum_{i=1}^n \sigma_i^2}.
$$
Note that this corresponds to taking the expectation of the numerator and denominator individually.

Using Jensen's inequality I can show
\begin{align}
\mathbb E\left[
\frac{x^2}{x^2 + z}
\right]
&=
\mathbb E\left[
\mathbb E\left[
\frac{x^2}{x^2 + z}
\mid x
\right]
\right]
\\&\ge
\mathbb E\left[
\frac{x^2}{x^2 + \mathbb E[z]}
\right]
\\&\approx
\frac{1}{1 + \sqrt{\mathbb E[z]} + \mathbb E[z]}.
\end{align}
However, what I would need to be true is
$
\mathbb E\left[
\frac{x^2}{x^2 + z}
\right] \ge \frac{1}{1 + \mathbb E[z]}
$, and that certainly doesn't hold in general.
In particular it seems I need to somehow use that it's the smallest $\sigma_n$ that's in the numerator. The equivalent result with an arbitrary $\sigma_i$ doesn't seem to be true in general.
It's also interesting to notice that in the simple case $n=2$ we get
$$
\mathbb E\left[
\frac{\sigma_2^2 x_2^2}{\sigma_1^2 x_1^2 + \sigma_2^2 x_2^2}
\right]
= \frac{\sigma_2}{\sigma_1 + \sigma_2}.
$$
(At least Mathematica says this is true, I'd be interested in knowing a proof.)
Though that definitely doesn't hold for $n > 2$.
I suppose the equation $\sum_{i=1}^n \sigma_i^2 x_i^2 = 1$ corresponds to integrating over an ellipse, but I haven't found a nice geometric way to make use of that.

I tried something else.
In the case $\sigma_1 = 1$ and $\sigma_2=\sigma_3$, Mathematica can evaluate the expectation as
$$
\frac{x_1^2}{x_1^2 + \sigma_2^2 (x_2^2 + x_3^2)}
=
\frac{1}{1-\sigma_2^2}+\frac{\sigma_2 \sinh ^{-1}\left(\sqrt{\sigma_2^2-1}\right)}{\left(\sigma_2^2-1\right)^{3/2}}.
$$
As expected this is below $1/(1+2\sigma_2^2)$ for $\sigma_2 > 1$:

Mathematica even finds an expression for the general case $E[\frac{x_n^2}{x_n^2+a \chi^2}]$ where $\chi^2$ is Chi-squared distributed with $n-1$ degrees of freedom.
So maybe there's a proof works by "evening out" the larger $\sigma$ values...
The bound with chi-squared isn't particularly pretty though...
A statement equivalent to my inequality is that
$$
\mathbb E\left[
\frac{x_n^2}{\sum_{i=1}^n p_i x_i^2}
\right]
\ge E\left[
\frac{x_n^2}{\frac{1}{n} \sum_{i=1}^n x_i^2}
\right],
$$
where $\sum_i p_i=1$, and $p_1 \ge p_2 \ge \dots \ge p_n \ge 0$.
Since $E\left[
\frac{x_n^2}{\sum_{i=1}^n x_i^2}
\right]=\frac1n$ by symmetry.
It might even be that all of this is true independent of $x_i$ being Gaussian, as long as they are IID.
 A: Let $p_i = \sigma_i^2/\sigma_n^2, \, i = 1, 2, \cdots, n-1$. Then
$p_1 \ge p_2 \ge \cdots \ge p_{n-1} \ge 1$. We need to prove that
$$\mathbb{E}\left[\frac{x_n^2}{\sum_{i=1}^{n-1} p_i x_i^2 + x_n^2} \right] \ge \frac{1}{\sum_{i=1}^{n-1}p_i + 1}.$$
Using the identity ($q > 0$)
$$
\frac{1}{q} = \int_0^\infty \mathrm{e}^{-qt}\, \mathrm{d} t,
$$
we have
\begin{align*}
 \mathbb{E}\left[\frac{x_n^2}{\sum_{i=1}^{n-1} p_i x_i^2 + x_n^2} \right]
 &= \mathbb{E}\left[ \int_0^\infty x_n^2 \mathrm{e}^{-t(\sum_{i=1}^{n-1} p_i x_i^2 + x_n^2)}\,\mathrm{d}t \right]\\
 &= \int_0^\infty \mathbb{E}[x_n^2\mathrm{e}^{-tx_n^2}] \prod_{i=1}^{n-1} \mathbb{E}[\mathrm{e}^{-tp_i x_i^2}] \,\mathrm{d} t\\
 &= \int_0^\infty \frac{1}{(2t+1)^{3/2}}\prod_{i=1}^{n-1} \frac{1}{\sqrt{2p_it + 1}} \,\mathrm{d} t \tag{1}\\
 &\ge \int_0^\infty \left(\frac{2(s + 3)t + n+2}{n+2}\right)^{-\frac{n+2}{2}}\,\mathrm{d} t \tag{2}\\
 &= \frac{n + 2}{n(s + 3)}\\
 &\ge \frac{1}{s + 1} \tag{3}
\end{align*}
where $s = p_1 + p_2 + \cdots + p_{n-1}$.
Explanations:
(1): We have used
$\mathbb{E}[x_n^2\mathrm{e}^{-tx_n^2}]
= \int_{-\infty}^\infty x^2 \mathrm{e}^{-tx^2}\cdot \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-x^2/2}\, \mathrm{d} x = \frac{1}{(2t + 1)^{3/2}}$
and $\mathbb{E}[\mathrm{e}^{-tp_i x_i^2}]
= \int_{-\infty}^\infty \mathrm{e}^{-tp_ix^2}\frac{1}{\sqrt{2\pi}} \mathrm{e}^{-x^2/2} \,\mathrm{d} x = \frac{1}{\sqrt{2p_i t + 1}}$;
(2): We have used AM-GM inequality;
(3): We have used
$\frac{n + 2}{n(s + 3)} - \frac{1}{s + 1} = \frac{2(s + 1 - n)}{n(s + 3)(s + 1)} \ge 0$ (using $s \ge n - 1$).
We are done.
A: Ok, I have a quite general proof now. It works for all IID sets of variables, not only Gaussian.
Let $a_1\dots,a_n\ge 0$ be IID random variables and assume some values $p_i \ge 0$ st. $\sum_i p_i=1$.
In particular assume one value $p_n \le 1/n$.
We will show
$$
E\left[\frac{a_n}{\sum_i p_i a_i}\right] \ge 1,
$$
which solves the original problem letting $p_i=\frac{\sigma_i^2}{\sum_j \sigma_j^2}$ and $a_i=x_i^2$.
Since $p_n$ is the smallest $p_i$, in particular it must be at most $1/n$.
Reduction from n variables to two:
Since the $a_i$ are IID, it doesn't matter if we permute some of them, in particular,
\begin{align}
E\left[\frac{a_n}{\sum_i p_i a_i}\right]
&=
E_a\left[E_{\pi}\left[\frac{a_n}{p_n a_n + \sum_i p_i a_{\pi_i}}\right]\right]
\\&\ge
E_a\left[\frac{a_n}{E_{\pi}\left[p_n a_n + \sum_{i<n} p_i a_{\pi_i}\right]}\right]
\\&=
E_a\left[\frac{a_n}{p_n a_n + \sum_{i<n} p_i ( \frac{1}{n-1}\!\sum_{j<n} a_j)}\right]
\\&=
E_a\left[\frac{a_n}{p_n a_n + (1-p_n)\sum_{i<n} \frac{a_i}{n-1}}\right],
\end{align}
Using Jensen's inequality on the convex function $1/x$.
Now, since $p_n \le 1/n$, we have $1-p_n\ge p_n$, so we have reduced to the case of just two variables.
Two variable case:
Maybe this case should be called something else, since as Drew Brady pointed out the two variables in this case are not identically distributed. Yet, we can still use Jensen in a similar way to before.
Define $a = \sum_{i=1}^n a_i$.
By permuting $a_n$ with the other variables, we get:
\begin{align}
E_a\left[
\frac{a_n}{p_n a_n + (1-p_n)\sum_{i<n} \frac{a_i}{n-1}}
\right]
&=
E_a\left[\frac{a_n}{p_n a_n + \frac{1-p_n}{n-1}(a-a_n)}\right]
\\&=
E_a\left[
\frac{1}{n}\sum_{i=1}^n
\frac{a_i}{p_n a_i + \frac{1-p_n}{n-1}(a-a_i)}\right]
\\&=
E_a\left[
\frac{1}{n}\sum_{i=1}^n
\frac{a_i/a}{\frac{1-p_n}{n-1} - (\frac{1-p_n}{n-1}-p_n) a_i/a}\right]
\\&=
E_a\left[
\frac{1}{n}\sum_{i=1}^n
\phi(a_i/a)
\right],
\end{align}
where
$$
\phi(q_i)
= \frac{q_i}{\frac{1-p_n}{n-1} - (\frac{1-p_n}{n-1}-p_n) q_i}
$$
is convex whenever
$\frac{1-p_n}{n-1} \big/ (\frac{1-p_n}{n-1}-p_n) = \frac{1-p}{1-np} \ge 1$,
which is true when $0\le p_n \le 1/n$.
That means we can use Jensen's again:
$$
\frac{1}{n}\sum_{i=1}^n
\phi(a_i/a)
\ge 
\phi\left(\frac{1}{n}\sum_i \frac{a_i}{a}\right)
=
\phi\left(\frac{1}{n}\right)
=
1,
$$
which is what we wanted to show.
