How to prove that $f(X)=\frac{\det(X)}{\mathrm{tr}(X)}$ is log-concave? How to prove that $f(X)=\frac{\det(X)}{\mathrm{tr}(X)}$, $X\in S^n_{++}$ is log-concave, i.e., $\log(f(X))$ is concave? The symbol $S^n_{++}$ denotes all $n\times n$ positive matrices.
I want to imitate the proof of that $\log(\det(X))$ is concave, but I can't tell the relation of eigenvalues of $Y$ and $X^{-1/2}YX^{-1/2}$.
 A: This is not true if you remove the positive-definiteness requirement:  $$f\left(\begin{bmatrix}2&0\\0&1\end{bmatrix}\right)=\frac{2}{3}$$
$$f\left(\begin{bmatrix}-2&0\\0&1\end{bmatrix}\right)=\frac{-2}{-1}=2$$
$$f\left(\frac{1}{2}\begin{bmatrix}2&0\\0&1\end{bmatrix}+\frac{1}{2}\begin{bmatrix}-2&0\\0&1\end{bmatrix}\right)=f\left(\begin{bmatrix}0&0\\0&1\end{bmatrix}\right)=\frac{0}{1}\ngeq\sqrt{\frac{2}{3}\cdot2}$$
(That requirement was missing from an earlier version of the question.)
A: Let $A$ be positive definite and $H$ be any Hermitian matrix. When $t\in\mathbb R$ is sufficiently small,
\begin{aligned}
\frac{d}{dt}\log\left(\frac{\det(A+tH)}{\operatorname{tr}(A+tH)}\right)
&=\frac{\frac{d}{dt}\det(A+tH)}{\det(A+tH)}-
\frac{\frac{d}{dt}\operatorname{tr}(A+tH)}{\operatorname{tr}(A+tH)}\\
&=\operatorname{tr}\left((A+tH)^{-1}H\right)-
\frac{\operatorname{tr}(H)}{\operatorname{tr}(A+tH)}.\\
\end{aligned}
Therefore
\begin{aligned}
&\left.\frac{d^2}{dt^2}\right|_{t=0}\log\left(\frac{\det(A+tH)}{\operatorname{tr}(A+tH)}\right)\\
&=\left[\operatorname{tr}\left(-(A+tH)^{-1}H(A+tH)^{-1}H\right)
+\frac{\operatorname{tr}(H)^2}{\operatorname{tr}(A+tH)^2}\right]_{t=0}\\
&=-\operatorname{tr}\left(A^{-1}HA^{-1}H\right)+\frac{\operatorname{tr}(H)^2}{\operatorname{tr}(A)^2}\\
&=\frac{-\operatorname{tr}(A)^2\operatorname{tr}(A^{-1}HA^{-1}H)+\operatorname{tr}(H)^2}{\operatorname{tr}(A)^2}\\
&\le\frac{-\operatorname{tr}(A^\color{red}{2})\operatorname{tr}(A^{-1}HA^{-1}H)+\operatorname{tr}(H)^2}{\operatorname{tr}(A)^2}\quad\text{(because $A\succ0$)}\\
&=\frac{-\operatorname{tr}(A^2)\operatorname{tr}\left((A^{-1/2}HA^{-1/2})^2\right)+\operatorname{tr}\left(A(A^{-1/2}HA^{-1/2})\right)^2}{\operatorname{tr}(A)^2}\\
&=\frac{-\|A\|_F^2\,\|A^{-1/2}HA^{-1/2})^2\|_F^2+\langle A,\,A^{-1/2}HA^{-1/2}\rangle_F^2}{\operatorname{tr}(A)^2}
\le0.
\end{aligned}
A: Here is a proof if one of the matrices is $1$. Call the eigenvalues of the other matrix $\lambda_i$. Then we want to prove
$$\frac{\prod \left(\frac{\lambda_i + 1}2 \right)^2}{\left(\sum \frac{\lambda_i + 1}2\right)^2} \geq \frac{\prod \lambda_i}{n\sum \lambda_i} \,.$$
Bring products to one side and sums to the other:
$$\prod \left(\frac{\lambda_i + 1}2 \right)\left(\frac{\lambda_i^{-1} + 1}2 \right) \geq \frac{\left( \sum (\lambda_i + 1) \right)^2}{4n\sum \lambda_i} \,.$$
We have $\left(\frac{\lambda_i + 1}2 \right)\left(\frac{\lambda_i^{-1} + 1}2 \right) = 1 + \frac{(\lambda_i-1)^2}{4\lambda_i}$ so this becomes
$$\prod \left(1 + \frac{(\lambda_i-1)^2}{4 \lambda_i} \right) \geq 1 +  \frac{\left( \sum (\lambda_i - 1) \right)^2}{4 n\sum \lambda_i} \,.$$
By throwing away positive terms and Cauchy-Schwarz in Engel form (Titu's lemma), the LHS is at least
$$\begin{align*}
1 + \sum \frac{(\lambda_i-1)^2}{4 \lambda_i}
&\geq 1 + \frac{(\sum (\lambda_i-1))^2}{\sum 4 \lambda_i}
\end{align*}
$$
and this is bigger than the RHS as $n \geq 1$. (This last step feels strange but seems correct?)
