I am reading a paper where it says the following:
Consider a sequence of $d\times d$ positive semidefinite matrices $X_1, \ldots, X_t$ with $\max_t tr(X_t) \leq 1$ and define $M_0 = \lambda I_{d\times d}, \ldots, M_t = M_{t-1} + X_t$. Then by concavity of the $\log \det (\cdot)$ function, we have
$$\log (\det (M_{t-1})) \leq \log(\det(M_t)) + tr(M_t^{-1}(M_{t-1} - M_t))$$
I am not sure how to derive this using concavity of $\log \det (\cdot)$ function and where does trace pops up from.
Let $P=M_t^{-1/2}M_{t-1}M_t^{-1/2}$, so that $0\prec P\preceq I$. The inequality in question can be rewritten as $$ \log\det(M_t^{-1/2}M_{t-1}M_t^{-1/2}) \le\operatorname{tr}(M_t^{-1/2}(M_{t-1}-M_t)M_t^{-1/2}). $$ That is, $$ \log\det(P)\le\operatorname{tr}(P-I).\tag{1} $$ If we denote by $\lambda_1,\ldots,\lambda_n$ the eigenvalues of $P$, then $(1)$ is equivalent to $\sum_i\log(\lambda_i)\le\sum_i(\lambda_i-1)$. Hence it can be immediately proved by noting that $\log(x)\le x-1$ for all $x>0$. Yet we shall explore how the authors of the paper might intend to use the log-concavity of the determinant function.
Let $A(s)=sP+(1-s)I$ and $f(s)=\log\det A(s)$. By Jacobi's formula, $$ f'(s) =\frac{1}{\det A(s)}\frac{d\det A(s)}{ds} =\operatorname{tr}\left(A(s)^{-1}A'(s)\right). $$ Since $f$ is the composition of the affine function $A$ and the concave function $\log\det$, it is concave. Therefore $f(y)-f(x)\le f'(x)(y-x)$ and we obtain \begin{aligned} \log\det(P) &=f(1)-f(0)\\ &\le f'(0)(1-0)\\ &=\operatorname{tr}\left(A(0)^{-1}A'(0)\right)\\ &=\operatorname{tr}(P-I). \end{aligned}