Given a set of projectors $\{P_i\}$ acting on a space $\mathcal H_S$, let $\Phi$ be the LCPT map defined by $$\Phi(\rho)=\sum_i P_i\rho P_i.$$ The goal is to show that $S(\Phi(\rho))\ge S(\rho)$. The Klein inequality $$S(\rho\Vert\Phi(\rho))=\text{tr}[\rho\log\rho]-\text{tr}[\rho\log\Phi(\rho)]\ge 0 $$ already implies $-\text{tr}[\rho\log\Phi(\rho)]\ge S(\rho)$, so all that is left is to show that $$S(\Phi(\rho))\ge-\text{tr}[\rho\log\Phi(\rho)].$$
So, using the spectral decomposition of $\rho$ I obtained $$\rho\log\Phi(\rho)=\rho\sum_{ij}\log(p_j)P_i|j\rangle\langle j|P_i=\sum_i\rho P_i\sum_j\log(p_j)|j\rangle\langle j|P_i $$ and by applying the cyclic property of the trace $$-\text{tr}[\rho\log\Phi(\rho)]=-\text{tr}\left[\sum_iP_i\rho P_i\sum_j\log(p_j)|j\rangle\langle j|\right]=-\text{tr}[\Phi(\rho)\log\rho].$$ However I realized something disquieting: the Klein inequality should also hold for $S(\Phi(\rho)\Vert \rho)$, implying $$S(\Phi(\rho))\le -\text{tr}[\Phi(\rho)\log\rho]=-\text{tr}[\rho\log\Phi(\rho)] $$ so either something has gone wrong with my calculations and the proof is done in a completely different way, or $S(\Phi(\rho))=-\text{tr}[\rho\log\Phi(\rho)]$ and I'm not seeing why. Either way, I'd appreciate some clarifying help!