A trace inequality for four orthogonal projections

Let $P,Q,R,S\in M_K(\mathbb C)$ be orthogonal projections satisfying

$$P\perp Q,\quad R\perp S,\quad \operatorname{Im}(P)\cap\operatorname{Im}(R)=\{0\},\quad \operatorname{Im}(Q)\cap\operatorname{Im}(S)=\{0\},$$ $$\operatorname{rank}(P)+\operatorname{rank}(Q)=\operatorname{rank}(R)+\operatorname{rank}(S),\quad \operatorname{rank}(P)+\operatorname{rank}(R)=\operatorname{rank}(Q)+\operatorname{rank}(S).$$

Define orthogonal projections using Moore–Penrose inverses:

$$P'=(P+R)^{-1/2}P(P+R)^{-1/2},\quad Q'=(Q+S)^{-1/2}Q(Q+S)^{-1/2},$$ $$R'=(P+R)^{-1/2}R(P+R)^{-1/2},\quad S'=(Q+S)^{-1/2}S(Q+S)^{-1/2}.$$

Projection trace inequality conjecture. Under these hypotheses,

$$\operatorname{Tr}(PR)+\operatorname{Tr}(QS)\geq\operatorname{Tr}(P'Q')+\operatorname{Tr}(R'S').$$

This is presented as the $N=2$ specialization of the preceding $F_p$ inequality and as a step toward proving convergence of the orthogonalization algorithm.