Let $\{V_i\}_{i=1}^N$ be a set of $n\times m$, $n\geq m$, real matrices of full column rank and let $X=X^\top\in\mathbb{R}^{n\times n}$ be a positive definite trace-one matrix. Moreover, let $A^{1/2}=(A^{1/2})^\top$ denote the (unique) symmetric square root of a positive semi-definite matrix $A$, $\mathrm{tr}\, A$ the trace of $A$, and $\log A$ the matrix logarithm of $A$.

Question: Does the inequality $$\tag{$1$} \label{1} \sum_{i=1}^N \mathrm{tr}\log \left(\frac{1}{mN}\sum_{j=1}^N V_i^\top X^{1/2}V_j(V_j^\top X V_j)^{-1} V_j^\top X^{1/2}V_i\right) - \sum_{i=1}^N \mathrm{tr} \log \left(V_i^\top X V_i\right) \ge 0 $$ hold true?

**My (very little) progress so far.**
Using the fact that $\mathrm{tr}\log Y+\mathrm{tr}\log Z=\mathrm{tr}\log YZ$ and $-\log\, Y=\log Y^{-1}$ for positive definite $Y$, $Z$, we can rewrite \eqref{1} as
$$\tag{$2$}
\sum_{i=1}^N \mathrm{tr}\log \left(\frac{1}{mN}\sum_{j=1}^N V_i^\top X^{1/2}V_j(V_j^\top X V_j)^{-1} V_j^\top X^{1/2}V_i\left(V_i^\top X V_i\right)^{-1}\right) \ge 0.
$$
Also, exploiting the formula $\mathrm{tr}\log Y =\log \det Y$, we get
$$\tag{$3$}
\sum_{i=1}^N \log\det \left(\frac{1}{mN}\sum_{j=1}^N V_i^\top X^{1/2}V_j(V_j^\top X V_j)^{-1} V_j^\top X^{1/2}V_i\left(V_i^\top X V_i\right)^{-1}\right) \ge 0,
$$
or, equivalently,
$$\tag{$3$}
\prod_{i=1}^N \det \left(\frac{1}{mN}\sum_{j=1}^N V_i^\top X^{1/2}V_j(V_j^\top X V_j)^{-1} V_j^\top X^{1/2}V_i\left(V_i^\top X V_i\right)^{-1}\right) \ge 1.
$$
I'm stuck at this point. I believe that the problem can be further simplified to some well-known determinant inequality. However, at the moment, I don't know how to proceed.

**Note 1.** This question can be seen as a generalization of this problem. Indeed, for $m=1$, \eqref{1} reduces exactly to the case treated in that question, which has been proved to be true.

**Note 2.** Numerical simulations suggest that \eqref{1} is very likely to be true.

**Edit.** I edited the inequality by replacing the factor $1/N$ in the first logarithm by $1/(mN)$. This is a somewhat "stronger" version which I believe is the "right" generalization of the aforementioned problem.