= Solution
The <quantum relative entropy> is $S(\rho\|\sigma)=\operatorname{Tr}\rho(\log_2\rho-\log_2\sigma)$ when the <support of a positive operator> $\rho$ is contained in that of $\sigma$, and $+\infty$ otherwise. Write spectral decompositions $\rho=\sum_i r_i|i\rangle\langle i|$ and $\sigma=\sum_j t_j|u_j\rangle\langle u_j|$, and set $q_i=\langle i|\sigma|i\rangle$.
The weights $w_{ij}=|\langle i|u_j\rangle|^2$ sum to one. Concavity of the scalar logarithm gives the <diagonal logarithm concavity bound>
$$
\langle i|\log_2\sigma|i\rangle
=\sum_jw_{ij}\log_2t_j
\leq\log_2\left(\sum_jw_{ij}t_j\right)=\log_2q_i.
$$
For $r_i>0$, support inclusion removes overlaps with zero eigenvalues of $\sigma$; alternatively take a positive regularization and pass to the limit. The two diagonal vectors $r$ and $q$ are probability distributions. Therefore
$$
\boxed{S(\rho\|\sigma)\geq\sum_i r_i\log_2\frac{r_i}{q_i}
=D(r\|q)\geq0.}
$$
This proves <nonnegativity of quantum relative entropy> without assuming the states commute. Equality in the strict logarithmic concavity condition makes each occupied $|i\rangle$ an eigenvector of $\sigma$; classical equality gives $q_i=r_i$ and no weight outside the support, hence equality occurs exactly for $\rho=\sigma$.
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