Concavity of $\ln \det$ I'm trying to prove that function $A \mapsto \ln\det A$ is concave on the set of symmetric definite positive matrices $\mathcal{S}_d^{++}(\mathbb{R})$ but without using the lemma that allows me to restrict myself to a line.
Hence, I'm trying to prove that $\forall t \in [0, 1]$, $\forall \Gamma, \Sigma  \in \mathcal{S}_d^{++}(\mathbb{R})$ we have
$$\ln\det(t\Gamma + (1 - t)\Sigma) \geq t \ln\det \Gamma + (1 - t)\ln\det\Sigma$$ i.e. $$\det(t\Gamma + (1 - t)\Sigma) \geq (\det \Gamma)^t (\det \Sigma)^{1 - t}$$
If $\Gamma$ and $\Sigma$ commute then they share a common basis of eigenvectors and the inequality reduces to proving
$$\displaystyle \prod_{i = 1}^d (t \gamma_i + (1 - t)\sigma_i) \geq \left(\prod_{i = 1}^d \gamma_i\right)^t \left(\prod_{i = 1}^d \sigma_i\right)^{1 - t}$$
which is a corollary of inequality $\forall u, v > 0$, $\forall p, q$ conjugates, $uv \leq \frac{u^p}p + \frac{v^q}q$. However, if they do not commute I'm lost.
 A: For any two positive definite matrices $A$ and $B$, we want to prove that for every $t\in[0,1]$,
$$
\log\det\left(tA+(1-t)B\right)
\ge t\log\det(A)+(1-t)\log\det(B).
$$
Let $C=A^{-1/2}BA^{-1/2}$. The LHS above is equal to
$$
\log\det\left(A^{1/2}\left(tI+(1-t)C\right)A^{1/2}\right)
=\log\det(A)+\log\det\left(tI+(1-t)C\right)
$$
while the RHS is equal to
$$
t\log\det(A)+(1-t)\log\det(A^{1/2}CA^{1/2})
=\log\det(A)+(1-t)\log\det(C).
$$
Therefore the inequality is equivalent to
$$
\log\det\left(tI+(1-t)C\right)
\ge(1-t)\log\det(C),
$$
which is true because by the concavity of the logarithm function on the positive real line, $\log(t+(1-t)\lambda)
\ge(1-t)\log(\lambda)$ for every eigenvalue $\lambda$ of $C$.
A: It's enough to show that
$$\det \left(\frac{A^2 + B^2}{2}\right) \ge \det A B$$
if $A$, $B$ are positive definite matrices. Now, if $A$, $B$ commute then we have $A^2 + B^2 - 2 AB = (A-B)^2 \succeq 0$, so we have the inequality $\frac{A^2+B^2}{2}\succeq AB$, so the corresponding inequality for the determinants. But we can still reduce to one of the matrices being $I$, using the idea of @user1551.  Indeed we have
$$\det\frac{A^2+B^2}{2} = \det A \det \frac{I+A^{-1} B^2 A^{-1}}{2} \det A = \det A^2 \cdot \det \frac{I+A^{-1} B^2 A^{-1}}{2}$$
while
$$\det A B = \det A^2 \cdot \det \sqrt{A^{-1} B^2 A^{-1}}$$
Now we have indeed
$$\frac{I + A^{-1} B^2 A^{-1}}{2} \succeq \sqrt{A^{-1} B^2 A^{-1} }$$
and so the inequality for determinants.
