Does it hold that $tr(ABC)\le tr (AB)$?

Question

$A,B$ and $C$ are all real orthogonal projections. So for each $X\in \{A,B,C\}$ it holds that $X^2=X=X^*.$ Does it hold that $$tr(ABC)\le tr (AB)?$$

If no, for what special cases could it hold?

Answer 1

I suppose you are talking about orthogonal projections in a Euclidean space. The inequality already fails when $A,B,C$ are rank-one matrices. For instance, if you pick two unit vectors $u$ and $v$ that are almost orthogonal to each other and set $A=uu^T$ and $B=vv^T$, then $\operatorname{tr}(AB)=(u^Tv)^2$. Now, if $C=ww^T$ for some unit vector $w$ that is somewhat halfway between $u$ and $v$, then $\operatorname{tr}(ABC)=(u^Tv)(v^Tw)(w^Tu)$ should have about the same order of magnitude as $u^Tv$ and hence it should be greater than $\operatorname{tr}(AB)$. For a concrete counterexample, take $u=(1,0)^T,\,v=(\sqrt{1-0.9^2},0.9)$ and $w=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})^T$. Numerically we have $$ \operatorname{tr}(ABC)=0.2912>0.19=\operatorname{tr}(AB). $$