If $P_1$ and $P_2$ are orthogonal projectors, then $\mathrm{tr}(P_1 P_2) \leq \mathrm{rk}(P_1 P_2).$ 
Prove or provide a counterexample:  If $P_1$ and $P_2$ are orthogonal projectors, then $\mathrm{tr}(P_1 P_2) \leq \mathrm{rk}(P_1 P_2),$ where $\mathrm{tr}$ denoted the trace and $\mathrm{rk}$ the rank.

Some attempt. Since $P_1$ is an orthogonal projector, $P_1 = P_1^\intercal = P_1^2.$ We can orthogonally diagonalise
$$
P_1 = x_1 x_1^\intercal + \ldots + x_k x_k^\intercal = XX^\intercal,
$$
where $X = [x_1, \ldots, x_k]$ and the $x_i$ are orthonormal and they span the image of $P_1.$ Then
$$
\mathrm{tr}(P_1 P_2) = \mathrm{tr}(X^\intercal P_2 X) = \sum_{i = 1}^k x_i^\intercal P_2 x_i.
$$
The right hand side is $\leq k = \mathrm{rk}(P_1)$ since $x_i^\intercal P_2 x_i = \|P_2 x_i\|^2 \leq \|x_i\|^2 = 1.$ This, of course, does not prove the result.
Partial solution. If $P_1P_2 = P_2 P_1,$ then $P_1 P_2$ is the orthogonal projector onto the intersection of the images of $P_1$ and $P_2$ (see my other question: decomposition of projectors). In this case, we want to prove $\mathrm{tr}(P) \leq \mathrm{rk}(P)$ which is true since all eigenvalues of $P$ are either zero or one and therefore the two numbers coincide.
Therefore, if there is a counterexample, it has to be one for which $P_1 P_2 \neq P_2 P_1$ and this means (proved in the same question linked above), with $V_i = \mathrm{Im}(P_i),$
$$
V_1 \neq (V_1 \cap V_2) \oplus (V_1 \cap V_2^\perp) \qquad \text{(equiv., } V_2 \neq (V_1 \cap V_2) \oplus (V_1^\perp \cap V_2 \text{).}
$$
 A: It suffices to assume that $P_1$ is an orthogonal projector and $P_2$ is any matrix that satisfies $0\preceq P_2\preceq I$. Then $0\preceq P_1P_2P_1\,(\preceq P_1IP_1=P_1^2=P_1)\preceq I$. Therefore $\operatorname{tr}(P_1P_2P_1)\le\operatorname{rank}(P_1P_2P_1)$ (if you don't see this, think about the unitary diagonalisation of $P_1P_2P_1$) and in turn
$$
\operatorname{tr}(P_1P_2)
=\operatorname{tr}(P_1^2P_2)
=\operatorname{tr}(P_1P_2P_1)
\le\operatorname{rank}(P_1P_2P_1)
\le\operatorname{rank}(P_1P_2).
$$
A: $\big \Vert P_1P_2\big \Vert_2\leq \big \Vert P_1\big \Vert_2\cdot \big \Vert P_2\big \Vert_2 \leq 1$
i.e. operator 2 norm is submultiplicative and a non-zero orthogonal projection has operator 2 norm of 1.  This means all eigenvalues of $(P_1P_2)$, which are necessarily real (why?), satisfy $\lambda_i\in [0,1]$
Then $(P_1P_2)$ has $n$ eigenvalues, $k$ of which are non-zero
$ \text{trace}\big(P_1P_2\big) =\sum_{i=1}^n \lambda_i= \sum_{i=1}^k \lambda_i\leq \sum_{i=1}^k 1 = \text{rank}\big(P_1P_2\big)$
since $P_1P_2$ is diagonalizable, ref: A and B are real, symmetric and positive semi-definite matrices of the same order; is AB diagonalizable?
note: the fact that $\big(P_1P_2\big)$ is diagonalizable is nice but technically not needed. If the matrix were defective its rank would be bounded below by the number of non-zero eigenvalues, $k$.
