inequality for lengths of projections inside a circle

Question

In the following picture, how can show that $c \leq a + b$?

In the picture, $x$, $y$ and $z$ are three vectors of equal length. We can split $x$ in a component parallel to $y$ and a component orthogonal to $y$. The value $a$ is the length of the orthogonal component. Similarly, we can split $z$ into components parallel and orthogonal to $y$, and $b$ is the length of the orthogonal component. Finally, we can split $z$ into components parallel and orthogonal to $x$, and $c$ is the length of this orthogonal component. (In the definition of $c$, we can swap the roles of $x$ and $z$, and get the same value $c$.)

While the picture is two-dimensional, I suspect the statement is true for $x,y,z \in \mathbb{R}^d$ for and $d\geq 2$. I would be happy with a proof for $d=2$, but even happier about a proof which can also cover the case $d>2$.

Answer 1

Letting $\theta$ and $\phi$ be the angles between $x$ and $y$ and between $y$ and $z$ respectively, and letting $x$, $y$, and $z$ have equal length $1$, we have: \begin{align*} a &= \sin \theta \\ b &= \sin \phi \\ c &= \sin (\theta + \phi) \end{align*} and $c < a + b$ follows from the fact that $\sin x$ is sublinear for $x \in [0, \pi]$. (Alternatively, note that $c = \sin (\theta + \phi) = \sin \theta \cos \phi + \sin \phi \cos \theta < a+b$ because $\cos \theta$ and $\cos \phi$ are each less than $1$.)