So I know that $\|P(x)+P(y)\|\leq \|x+y\|$ where $P$ denote the projection of a vector to a convex set $C$ (I assume that $C$ is symmetric). How about the result of three terms? i.e. does $\|P(x)+P(y)+P(z)\|\leq \|x+y+z\|$ hold?
I read the proof of the two-term result which use Cauchy inequality and $\langle x-P(x),c-P(x) \rangle \leq 0$ for any $c\in C$, but this seems hard to be used in the three-term case.
I appreciate any advice, thank you!
I assume that the projection $P(x)$ of $x$ to $C$ is the closest to $x$ point of $C$. The three-term inequality can fail even in one-dimensional case for symmetric $C$. Indeed, let $C=[-1,1]$, $x=2$, $y=z=-1$. Then $x+y+z=0$, but $P(x)+P(y)+P(z)=1+(-1)+(-1)=-1$.