= Solution
\b[The concavity inequality is true, but its printed equality condition is false.] At $b=1$ equality always holds, even for distinct states; for example take $\rho_1=|0\rangle\langle0|$ and $\rho_2=|1\rangle\langle1|$. The correct statement on $0\le b\le1$ is
$$
\boxed{\text{Equality holds iff }b\in\{0,1\}\text{ or }\rho_1=\rho_2.}
$$
For $0<b<1$, entropy is therefore strictly concave.
We give the <matrix> inequality needed for the proof. Define the <quantum relative entropy> $D(\tau\|\sigma)=\operatorname{tr}\tau(\log_2\tau-\log_2\sigma)$ if the support of $\tau$ is contained in that of $\sigma$, and $+\infty$ otherwise. Diagonalize the two states with <eigenvalues> $r_i,s_j$, and put $q_{ij}=|\langle i|j\rangle|^2$. Each row and column of $q$ sums to one. The elementary scalar inequality
$$
u\ln(u/v)\ge u-v
$$
follows from $x\ln x-x+1\ge0$, with equality exactly at $x=1$; use limits at zero. Consequently
$$
D(\tau\|\sigma)=\frac1{\ln2}\sum_{ij}q_{ij}r_i\ln(r_i/s_j)
\ge\frac1{\ln2}\sum_{ij}q_{ij}(r_i-s_j)=0.
$$
Equality forces $r_i=s_j$ whenever $q_{ij}>0$. Hence every <eigenvector> $|i\rangle$ of $\tau$ is also a $\sigma$-<eigenvector> with <eigenvalue> $r_i$, so $\tau=\sigma$. Conversely identical states give zero relative entropy. This proves <nonnegativity of quantum relative entropy> and its equality condition, including states with zero <eigenvalues> under the support convention.
Now set $\rho=b\rho_1+(1-b)\rho_2$. For $0<b<1$ its support contains both component supports. Direct expansion gives
$$
S(\rho)-bS(\rho_1)-(1-b)S(\rho_2)
=bD(\rho_1\|\rho)+(1-b)D(\rho_2\|\rho)\ge0.
$$
Since both coefficients are positive, equality holds exactly when both component states equal $\rho$, equivalently $\rho_1=\rho_2$. At either endpoint the mixture has only one component and equality is automatic. This proves the corrected <strict concavity of Von Neumann entropy>.
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