Suppose we have Hermitian operators $P_1$ and $P_2$ on Hilbert space $\mathbb{H} \cong \mathbb{C}^n$, such that: $P_1^2=P_1$, $P_2^2=P_2$, $\{P_1, P_2\} = 0$. Let $x$ be a unit $L_2$-norm vector.
Show that $(x^TP_1x)^2 + (x^TP_2x)^2 \leq 1$
I can only see how one can show that the same inequality holds for $RHS=2$. That can be easily seen because of the fact that a projection of a vector is always "shorter" than the vector. But I'm confused about the $RHS=1$ case.
I think I should somehow use the restriction on the anticommutator, but I can't figure that out. It is easy to prove that in this case $P_1+P_2$ is also a projector but I can't see how this helps here.
In fact, the correct statement should be$$ x^TP_1x + x^TP_2x \leq 1,\quad\cdots(*). $$ (But your version is still true as we can see by squaring $(*)$.) This is called Bessel's inequality. To prove this, let $$ x= P_1x + P_2 x + P_3x,$$ where $P_3 = I-P_1-P_2$. We can see that $P_3$ is also a projection and $P_iP_j = O$ for $i\neq j$. Thus, we have $$ 1=x^Tx =\sum_{i=1}^3 x^TP_ix +\sum_{i\neq j}x^TP_iP_jx= \sum_{i=1}^3 x^TP_ix \geq x^TP_1x + x^TP_2x, $$ as desired.
$\textbf{EDIT:}$ I added short proof that if $P,Q$ are orthogonal projections such that $PQ+QP=0$, then $PQ=QP=0$.
Assume to the contrary that $PQ\neq 0$. Then there is $v$ such that $v=Qv$ and $PQv=Pv\neq 0$. By the assumption that $PQ=-QP$, we have $$ Pv = -QPv. $$ This implies that $Q$ has $-1$ as an eigenvalue. This leads to a contradiction.