Let $\{v\_i\}\_{i=1}^N$ be a set of $n$-dimensional real vectors and let $X=X^\top\in\mathbb{R}^{n\times n}$ be a positive definite trace-one matrix. I would like to prove (or disprove) the following inequality:
$$
\sum\_{i=1}^N \log \left(\frac{1}{N}\sum\_{j=1}^N \frac{v\_i^\top X^{1/2}v\_j v\_j^\top X^{1/2}v\_i}{v\_j^\top X v\_j}\right) - \sum\_{i=1}^N \log v\_i^\top X v\_i \geq 0,\qquad (\*)
$$
where $X^{1/2}=(X^{1/2})^\top$ denotes the principal square root of $X$. Note that the previous inequality holds for $N=1$. Indeed,
$$
\log \frac{\left(v\_1^\top X^{1/2}v\_1\right)^2}{v\_1^\top X v\_1} - \log v\_1^\top X v\_1 = 2 \log\frac{v\_1^\top X^{1/2}v\_1}{v\_1^\top X v\_1}\geq 0,
$$
where the last step follows from the fact that, since $X$ is a trace-one matrix, $v^\top(X^{1/2}-X)v\geq 0$ for all $v\in\mathbb{R}^n$. I carried out a large series of numerical simulations for $N>1$ in order to find a counterexample to $(\*)$, but I couldn't find it. Hence my guess is that $(\*)$ is true.


Thanks for your help!


**Some comments.** Notice that $(\*)$ can be written as
$$
\sum\_{i=1}^N \log \left(\frac{1}{N}\sum\_{j=1}^N \frac{v\_i^\top X^{1/2}v\_j v\_j^\top X^{1/2}v\_i}{v\_j^\top X v\_jv\_i^\top X v\_i }\right) \geq 0.
$$
A lower bound to the LHS of previous inequality can be found by using the inequality $\log(x)\geq 1-1/x$, which yields,
$$
\sum\_{i=1}^N \log \left(\frac{1}{N}\sum\_{j=1}^N \frac{v\_i^\top X^{1/2}v\_j v\_j^\top X^{1/2}v\_i}{v\_j^\top X v\_jv\_i^\top X v\_i }\right) \geq N -N\sum\_{i=1}^N \left(\sum\_{j=1}^N \frac{v\_i^\top X^{1/2}v\_j v\_j^\top X^{1/2}v\_i}{v\_j^\top X v\_jv\_i^\top X v\_i }\right)^{-1}.
$$
Hence if we prove that
$$
\sum\_{i=1}^N\left(\sum\_{j=1}^N \frac{v\_i^\top X^{1/2}v\_j v\_j^\top X^{1/2}v\_i}{v\_j^\top X v\_jv\_i^\top X v\_i }\right)^{-1}\leq 1
$$
we are done. Numerical simulations suggest that the latter inequality holds.




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**EDIT.** A follow-up of this question has been posted [here](https://mathoverflow.net/questions/260747/a-generalized-log-inequality-for-positive-definite-trace-one-matrices).



A log inequality for positive definite trace-one matrices
The proof of the general case, in a strong form suggested in the end of OP.


Denote $X^{1/4} v\_i=u\_i$, $X^{1/2}=S$, then we have ${\rm tr}\,S^2=1$ and need to prove that 
$$
{\rm tr}\,S^2\geqslant 
\sum\_i \frac{(Su\_i,u\_i) (Su\_j,u\_j)}{\sum\_j (u\_i,u\_j)^2},
$$
then the very original inequality follows by applying AM-GM to $N$ summands in RHS.


By homogeneity in $u$'s we may suppose that $(Su\_i,u\_i)=1$ for all $i$. On the set of symmetric operators on $\mathbb{R}^n$ we have an inner product $(Y,Z)={\rm tr}\, YZ$, so ${\rm tr}\,S^2=\|S\|^2$. Our restriction on $S$ may be rewritten as $(S,u\_i\otimes u\_i)=1$, where we naturally identify $u\otimes u$ with the (rank at most 1) operator $x\rightarrow (u,x)u$. For operators $T\_i=u\_i\otimes u\_i$, $i=1,\dots,N$ we have $(u\_i,u\_j)^2=(T\_i,T\_j)$. That is, we get the following problem: given that $(T\_i,T\_j)\geqslant 0$ for all $i,j$ and $(S,T\_i)=1$ prove that 
$$
\|S\|^2 \geqslant \sum\_{i=1}^N (T\_i,T\_1+\dots+T\_N)^{-1}
$$
For $c\_i=(T\_i,\sum T\_j)$ we choose numbers $\mu\_1,\dots,\mu\_N$ such that $\sum 1/c\_i=(\sum \mu\_i)^2/\sum \mu\_i^2c\_i$, for example take $\mu\_i=1/c\_i$. Then 
$$
\|S\|^2\sum \mu\_i^2c\_i=\|S\|^2 \sum\_i \sum\_j \mu\_i^2 (T\_i,T\_j)\geqslant \|S\|^2\|\sum \mu\_iT\_i\|^2\geqslant (S,\sum \mu\_iT\_i)^2=(\sum \mu\_i)^2
$$
as desired (we have used that $\mu\_i^2 (T\_i,T\_j)+\mu\_j^2 (T\_j,T\_i)\geqslant 2\mu\_i\mu\_j (T\_i,T\_j)$ for all pairs $i\ne j$.)


