i'm gonna start answering the question. actually, i am exploring an answer.
so, first, think of this in terms of an Inner Product Space or Hilbert Space (but in this case there are no complex conjugates that those familiar with Hilbert Spaces might miss).
let $\mathbf{x}$ be the point or element in that space that is fully defined by the real and periodic function $x(t)$ where $x(t+2\pi)=x(t) \quad \forall t$. if i remember correctly Hilbert Spaces are also Normed Linear Spaces (Banach Spaces).
there's an inner product
$$ \langle \mathbf{x}, \mathbf{y} \rangle \triangleq \int\limits_{-\pi}^{\pi} x(t) y(t) \ dt$$
a norm
$$ \| \mathbf{x} \| = \sqrt{ \langle \mathbf{x}, \mathbf{x} \rangle } $$
$$ \| \mathbf{x} \|^2 = \langle \mathbf{x}, \mathbf{x} \rangle $$
which is the distance that $\mathbf{x}$ is from the $\mathbf{0}$ element.
there are distance metrics between two points. the more general distance function is
$$ \begin{align}
\| \mathbf{x}-\mathbf{y} \| & = \sqrt{\| \mathbf{x}-\mathbf{y} \|^2} \\
& = \sqrt{ \langle \mathbf{x}-\mathbf{y}, \mathbf{x}-\mathbf{y} \rangle } \\
& = \sqrt{ \langle \mathbf{x}, \mathbf{x}-\mathbf{y} \rangle - \langle \mathbf{y}, \mathbf{x}-\mathbf{y} \rangle } \\
& = \sqrt{ \langle \mathbf{x}, \mathbf{x} \rangle + \langle \mathbf{y}, \mathbf{y} \rangle - 2 \langle \mathbf{x}, \mathbf{y} \rangle } \\
& = \sqrt{ \| \mathbf{x} \|^2 + \| \mathbf{y} \|^2 - 2 \langle \mathbf{x}, \mathbf{y} \rangle } \\
\end{align}$$
what this means is that, as long as the norms of $\mathbf{x}$ and $\mathbf{y}$ remain constant, the distance between $\mathbf{x}$ and $\mathbf{y}$ is minimized when the inner product between the two is maximized. and vise versa.
so now let's generalize $\mathbf{x}$ or $\mathbf{y}$ or $\mathbf{z}$ a little more. the point $\mathbf{x}$ corresponds to the periodic $x(t)$ and $\mathbf{y}$ corresponds to $y(t)$. another point in this Hilbert space, we'll call $\mathbf{x}(\tau)$ corresponds to the periodic function $x(t+\tau)$, and naturally $\mathbf{y}(\tau)$ corresponds to $y(t+\tau)$. so as $\tau$ moves from $0$ to $2\pi$, $\mathbf{y}(\tau)$ traces out a curve (what we used to call a parametric line curve) in this Hilbert space that is a closed curve, because $\mathbf{y}(2\pi) = \mathbf{y}(0) = \mathbf{y}$.
note that no matter what real value $\tau$ is,
$$ \| \mathbf{y}(\tau) \| = \| \mathbf{y} \| . $$
which means that all of the points on this closed curve are equal distance from the origin $\mathbf{0}$ no matter what $\tau$ is. and the same is true for $\mathbf{x}(\tau)$ and $\mathbf{z}(\tau)$.
now, without loss of generality, i think we can assume that $\tau_{xy}$ and $\tau_{yz}$ are zero, for which we then have to prove that $\tau_{xz}=0$ ($\pm k 2 \pi$, which i'm not gonna worry about). that means that the distance that $\mathbf{x}(0)$ is from $\mathbf{y}(0)$ is less than the distance $\mathbf{x}(0)$ is from $\mathbf{y}(\tau)$ for any other $\tau \ne k 2 \pi$.
but this curve $\mathbf{x}(\tau)$ is parallel in some sense to the curve $\mathbf{y}(\tau)$ because we know that $\mathbf{x}(\tau_1)$ is closest to the point $\mathbf{y}(\tau_1)$ and $\mathbf{x}(\tau_2)$ is closest to the point $\mathbf{y}(\tau_2)$ for any values of $\tau_1$ and $\tau_2$ for the same reason that $\mathbf{x}(0)$ is closest to the point $\mathbf{y}(0)$.
the same can be said of the curves $\mathbf{y}(\tau)$ and $\mathbf{z}(\tau)$ because of the assumption $\tau_{yz}=0$ which means that the distance that $\mathbf{y}(0)$ is from $\mathbf{z}(0)$ is less than the distance $\mathbf{y}(0)$ is from $\mathbf{z}(\tau)$ for any other $\tau \ne k 2 \pi$.
so, somehow, in a formal manner, can't we say that if $\mathbf{y}(0)$ is the element of $\mathbf{y}(\tau)$ closest to $\mathbf{x}(0)$ and $\mathbf{z}(0)$ is the element of $\mathbf{z}(\tau)$ that is closest to $\mathbf{y}(0)$, doesn't this mean that $\mathbf{z}(0)$ is the element of $\mathbf{z}(\tau)$ closest to $\mathbf{x}(0)$??
can we make this into a decent proof?
as you can see, i am willing to pay some bounty to whomever can help me formalize this (or to persuasively disprove my reasoning above).