A log inequality for positive definite trace-one matrices 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.

EDIT. A follow-up of this question has been posted here.
 A: 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 $i\ne j$.)
A: Here is the proof for $N=2$. It contains also some, well, you may call it 'ideas', which theoretically may help in further cases. 
Here is a straightforward proof that for two-dimensional vectors $x=(x_1,x_2),y=(y_1,y_2)$ we have $$(x_1^2+x_2^2)(y_1^2+y_2^2)+(x_1y_1+x_2y_2)^2\geqslant \sqrt{(x_1^4+x_2^4)(y_1^4+y_2^4)}+x_1^2y_1^2+x_2^2y_2^2.$$
We use notations $\|x\|^2=\sum x_i^2$ for $x=(x_1,\dots,x_d)\in \mathbb R^d$, $(x,y)=\sum_{i=1}^d x_iy_i$ denotes the standard inner product.
Next, I need an easy 
Lemma. For vectors $p,q,a$ in $\mathbb{R}^d$ we have 
$$2\frac{(p,a)\cdot (q,a)}{\|a\|^2}\leqslant \|p\|\cdot \|q\|+(p,q)$$
Proof. At first, replacing $a$ to its projection onto 2-plane containing $x$, $y$ increases LHS and does not change RHS. So we may suppose that $p,q,a$ lie in a 2-plane. Then if we denote angles between $p,a$ by $\alpha$ and $q,a$ by $\beta$ we have to prove $2\cos \alpha \cos \beta\leqslant 1+\cos(\alpha\pm \beta)$, this follows from identity $2\cos \alpha \cos \beta=\cos(\alpha+\beta)+\cos(\alpha-\beta)$.
More involved observation is the following. Assume that $L$ is a subspace in $\mathbb{R}^d$ $T$ is a self-adjoint operator on $\mathbb{R}^d$. Then there exists a non-negative definite self-adjoint operator $S:L\rightarrow L$ such that $\|Su\|=\|Tu\|$ for any $u$ in $L$. 
Proof. Let $Q$ be any linear isometry from $T(L)$ to $L$ (if $T(L)$ has dimension less than that of $L$, $Q$ is defined on some extension of $T(L)$). Then $QT:L\rightarrow L$ and $\|QTu\|=\|Tu\|$ for $u\in L$. But $QT$ is not self-adjoint, well, take $S=\sqrt{(QT)^*QT}$.
Now come to your question for $N=2$. Denote $x=X^{1/4} v_1$, $y=X^{1/4}v_2$, $T=X^{1/4}$, ${\rm tr}\, T^4=1$. Your inequality may be rewritten as
$$
\left(\frac{\|x\|^4}{\|Tx\|^4}+\frac{(x,y)^2}{\|Tx\|^2 \|Ty\|^2}\right)
\left(\frac{\|y\|^4}{\|Ty\|^4}+\frac{(x,y)^2}{\|Tx\|^2 \|Ty\|^2}\right)\geqslant \frac 4{({\rm tr}\, T^4)^2},
$$
where last denominator is added for making it homogeneous in $T$. 
We check it in two steps: at first, prove it for $S$ instead $T$, next, prove that ${\rm tr}\, S^4\leqslant {\rm tr}\, T^4$. The second step is a standard application of variational principle. If $\alpha\geqslant \beta\geqslant 0$ are eigenvalues of $S$, $\lambda_1\geqslant \dots \geqslant \lambda_n>0$ are eigenvalues of $T$, we see that there exists non-zero vector $u\in L$ such that $\lambda_1 \|u\|\geqslant\|Tu\|=\|Su\|=\alpha\cdot \|u\|$. It follows that $\alpha\leqslant \lambda_1$. Next, we choose a non-zero vector $u$ which lies both in $L$ and in a subspace $W$ of codimension 1 generated by all eigenvectors of $T$ corresponded to $\lambda_2,\lambda_3,\dots,\lambda_n$. Then $\lambda_2 \|u\|\geqslant \|Tu\|=\|Su\|\geqslant \beta\cdot \|u\|$, thus $\lambda_2\geqslant \beta$. The claim ${\rm tr}\, S^4\leqslant {\rm tr}\, T^4$ follows. 
Now the first step. By Cauchy-Bunyakovsky-Schwarz Inequality we have 
$$
\left(\frac{\|x\|^4}{\|Sx\|^4}+\frac{(x,y)^2}{\|Sx\|^2 \|Sy\|^2}\right)
\left(\frac{\|y\|^4}{\|Sy\|^4}+\frac{(x,y)^2}{\|Sx\|^2 \|Sy\|^2}\right)\geqslant
\left(\frac{\|x\|^2\cdot \|y\|^2+(x,y)^2}{\|Sx\|^2 \|Sy\|^2}\right)^2,
$$
and it remains to prove that
$$
\frac{\|x\|^2\cdot \|y\|^2+(x,y)^2}{\|Sx\|^2 \|Sy\|^2}\geqslant \frac2{{\rm tr}\, S^4}.
$$
We may suppose that $S$ is diagonal with diagonal elements $\alpha,\beta$. Also denote $x=(x_1,x_2),y=(y_1,y_2)$. For $a=(k_1^2,k_2^2),b=(x_1^2,x_2^2),p=(k_1^2,k_2^2)$ our lemma says that 
$$
2\frac{\|Sx\|^2 \|Sy\|^2}{{\rm tr}\, S^4}\leqslant \sqrt{(x_1^4+x_2^4)(y_1^4+y_2^4)}+x_1^2y_1^2+x_2^2y_2^2, 
$$
and it remains to use out starting inequality.
