# inequality for lengths of projections inside a circle

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 $$d>2$$.

• Brief comment: "flattening" any arrangement of three vectors in $\mathbb{R}^3$ by rotating $x$ about $y$ until $x, y, z$ are coplanar will lengthen $c$ while keeping $a$ and $b$ the same, and any three vectors in $\mathbb{R}^d$ define a subspace of at most dimension $3$, so proving the two-dimensional case is sufficient. Commented Aug 20, 2019 at 16:57
• (Should read: "until $x, y, z$ are coplanar and $y$ is between $x$ and $z$"; sorry. And I'm not sure that in the three-dimensional case, both lines marked as $c$ have equal length.) Commented Aug 20, 2019 at 20:04

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 $$1$$.)
• I agree, this seems to work if $x$, $y$ and $z$ are arranged as in the picture. But does this really cover all possible arrangements? What if $z$ is between $x$ and $y$? What if $\theta+\phi > \pi$? Commented Aug 20, 2019 at 17:38
• Whether this works if either $\theta$ or $\phi$ exceeds $\pi/2$ depends on how you define signed lengths, I believe. Commented Aug 20, 2019 at 20:02
• $a$, $b$ and $c$ in the question are meant to be positive. So "signed lengths" may be part of the proof, but they are not part of the statement. Commented Aug 20, 2019 at 20:31