TITLE: A generalized log inequality for positive definite trace-one matrices
QUESTION [8 upvotes]: 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.

REPLY [3 votes]: Fedor's proof can be generalized :
Let $S = X^{1/2}$, $U_i = V_i S^{1/2}$ and $W_i = U_i (U_i^T S U_i)^{-1/2}$ .
Then $W_i^T S W_i = I_m$ where $I_m$ is the $m\times m$ identity .
Since $log(x) \ge 1 - 1/x$ it is enough to show that 
$$tr \sum_{i=1}^N \mu_i \le 1$$ 
where 
$$\mu_i = (\sum_{j=1}^N W_i^T W_j W_j^T W_i)^{-1}$$ .
Then 
$$tr \sum_{i=1}^N \mu_i = tr \sum_{i=1}^N \mu_i^2 \mu_i^{-1}$$
$$= tr [\sum_{i=1}^N \mu_i^2 \sum_{j=1}^N W_i^T W_j W_j^T W_i]$$
$$\ge tr \sum_{i=1}^N \sum_{j=1}^N W_i \mu_i W_i^T W_j \mu_j W_j^T$$
$$= tr [(\sum_{i=1}^N W_i \mu_i W_i^T)^2]$$
$$\ge [tr (S \sum_{i=1}^N W_i \mu_i W_i^T)]^2$$
$$= (tr \sum_{i=1}^N \mu_i)^2$$ ,
where I have used that
$$tr (\mu_i^2 W_i^T W_j W_j^T W_i) + tr (\mu_j^2 W_j^T W_i W_i^T W_j) \ge 2 tr (W_i \mu_i W_i^T W_j \mu_j W_j^T)$$ .
This follows from
$$0 \le tr[(W_i^T W_j \mu_j - \mu_i W_i^T W_j) (\mu_j W_j^T W_i - W_j^T W_i \mu_i)]$$ .