 Frobenius rank inequality

For compatible matrices $P,Q,R$, the inequality is $\mathrm{rank}(PQ)+\mathrm{rank}(QR)\le \mathrm{rank}(Q)+\mathrm{rank}(PQR)$. One proof uses the block matrix $M=\left(\begin{array}{cc}PQ& 0\\ Q& QR\end{array}\right)$. Subtract $P$ times its lower block row from its upper block row and then subtract its first block column times $R$ from its second. The result is $\left(\begin{array}{cc}0& -PQR\\ Q& 0\end{array}\right)$, with rank $\mathrm{rank}(Q)+\mathrm{rank}(PQR)$. On the other hand, choose independent columns of $PQ$ from the first block column and independent columns of $QR$ from the second. The corresponding columns of $M$ are independent: project a relation onto the upper block first, and then onto the lower block. Thus $\mathrm{rank}(M)\ge \mathrm{rank}(PQ)+\mathrm{rank}(QR)$.