How to show the inequality with the matrix fraction? I'm working on a proof where one step is to show the following inequality
$$
{\rm tr}\left[\left(\sum_{i=1}^{n}p_ix_ix_i^\top\sqrt{x_i^\top\mathbf{\Sigma}^{-2}x_i}\right)^{-1}\right]
\leq \sum_{i=1}^n p_i \sqrt{x_i^\top\mathbf{\Sigma}^{-2}x_i},
$$
where $x_i\in \mathbb{R}^{d\times 1}$, $\mathbf{\Sigma}=\sum_{i=1}^{n}p_ix_ix_i^\top$, and $p_i$'s are probabilities with $\sum_{i=1}^np_i=1$. It's easy to check that this inequality holds when $d=1$ by using Jensen inequality. I would like to ask how to show it for the high-dimensional case. Thank you very much!
 A: (Edit. The OP has clarified that the LHS of the inequality is the trace of a matrix inverse rather than the inverse of a matrix trace. So, this answer does not address the OP, but I'll leave it here if anyone find it useful. )
Your proof of the scalar case can be extended to higher dimensions. Since
$$
\sum_i \frac1d\ p_i x_i^T\Sigma^{-1}x_i
= \frac1d\operatorname{tr}\left(\Sigma^{-1} \sum_i p_ix_ix_i^T\right)
= \frac1d\operatorname{tr}(I_d)=1,
$$
the values $\frac1d p_i x_i^T\Sigma^{-1}x_i$ for $i=1,2,\ldots,n$ make up a probability vector. Therefore
\begin{aligned}
\left[\operatorname{tr}\left(\sum_{i=1}^{n}p_ix_ix_i^\top\sqrt{x_i^\top\Sigma^{-2}x_i}\right)\right]^{-1}
&=\left(\sum_{i=1}^{n}p_i\|x_i\|^2\|\Sigma^{-1}x_i\|\right)^{-1}\\
&\le\left(\sum_{i=1}^{n}p_i\|x_i\|\ (x_i^T\Sigma^{-1}x_i)\right)^{-1}\ \text{(Cauchy-Schwarz inequality)}\\
&=\frac1d\left(\sum_{i=1}^{n}\left(\frac1d\ p_ix_i^T\Sigma^{-1}x_i\right)\|x_i\|\right)^{-1}\\
&\le\frac1d\sum_{i=1}^{n}\left(\frac1d\ p_ix_i^T\Sigma^{-1}x_i\right)\frac1{\|x_i\|}\ \text{(Jensen's inequality)}\\
&\le\frac1d\sum_{i=1}^{n}\frac1d\ p_i\|x_i\|\|\Sigma^{-1}x_i\|\frac1{\|x_i\|}\ \text{(Cauchy-Schwarz inequality)}\\
&=\frac1{d^2}\sum_{i=1}^{n}p_i\|\Sigma^{-1}x_i\|\\
&=\frac1{d^2}\sum_{i=1}^n p_i \sqrt{x_i^\top\Sigma^{-2}x_i}\\
&\le\sum_{i=1}^n p_i \sqrt{x_i^\top\Sigma^{-2}x_i}.
\end{aligned}
