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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.

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.

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.

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.

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 previous inequality can be found 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.

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.