Prove that the Euclidean distance is no more than the spherical distance In an $n$-dimensional Euclidean space, for any unit vector $$\boldsymbol{y}=(y_1,y_2,\cdots, y_n)\in\mathbb{S}^{n-1}=\{\boldsymbol{x}\in\mathbb{R}^n:\|\boldsymbol{x}\|_2=1\},$$
we can express $\boldsymbol{y}$ in the spherical coordinate system via $\phi(\boldsymbol{y})=(\phi_1(\boldsymbol{y}), \phi_2(\boldsymbol{y}), \ldots, \phi_{n-1}(\boldsymbol{y}))^{\mathrm{T}}\in\mathbb{R}^{n-1}$ s.t.
$$
\begin{aligned}
y_1 &=\cos \phi_1 \\
y_2 &=\sin \phi_1 \cos \phi_2 \\
y_3 &=\sin \phi_1 \sin \phi_2 \cos \phi_3 \\
& ~\,\,\vdots \\
y_{n-1} &=\sin \phi_1 \cdots \sin \phi_{n-2} \cos \phi_{n-1} \\
y_{n} &=\sin \phi_1 \cdots \sin \phi_{n-2} \sin \phi_{n-1},
\end{aligned}
$$
where $0 \leq \phi_{n-1}<2 \pi$, and $0 \leq \phi_{i}\le \pi$, $\forall\,i=1,2,\ldots,n-2$.
My question is, if the Euclidean distance is no more than the spherical distance, i.e., if one has
$$
\boxed{\|\boldsymbol{x}-\boldsymbol{y}\|_2 \leq\|\phi(\boldsymbol{x})-\phi(\boldsymbol{y})\|_2}.
$$
I have been thinking this for several days, but am still unable to prove it. What is easy to show is that $\|\boldsymbol{x}-\boldsymbol{y}\|_2 \leq\|\phi(\boldsymbol{x})-\phi(\boldsymbol{y})\|_1$, which simply follows from the fact that the Euclidean distance is the shortest between any two points, and $|\phi_i(\boldsymbol{x})-\phi_i(\boldsymbol{y})|$ is exactly the length of the arc used to align $\boldsymbol{x}$ and $\boldsymbol{y}$ along the $i$-th spherical coordinate, which is longer than the corresponding chord.
If this is not true, then does it hold for any two close vectors $\boldsymbol{x}$ and $\boldsymbol{y}$ in the sense that $\underset{i=1,2,\ldots,n}{\max}|\phi_i(\boldsymbol{x})-\phi_i(\boldsymbol{y})|\le\delta$ for some small $\delta$?
 A: Yes, the Euclidean distance is no more than "the spherical distance" as defined in the question.
Christophe Leuridan's answer illustrates how this can be proved by "the mean value inequality for vector-to-vector maps",  such as proposition 2.7.6 of Lecture Notes on Multivariable Calculus. This answer provides a more elementary proof.

Let $n\ge2$, $\boldsymbol x, \boldsymbol y\in \Bbb S^{n-1}$. Suppose $\phi(\boldsymbol x)=(\alpha_1, \cdots, \alpha_n)$ and $\phi(\boldsymbol y)=(\beta_1,\cdots, \beta_n)$.
What we need to prove is the following claim, where the LHS of the inequality is $(
\|\boldsymbol{x}-\boldsymbol{y}\|_2)^2$ while the RHS is $(\|\phi(\boldsymbol{x})-\phi(\boldsymbol{y})\|_2)^2$.
Claim: we have
$$
\sum_{i=1}^{n-1}(\cos\alpha_i\prod_{j=1}^{i-1}\sin\alpha_j-\cos\beta_i\prod_{j=1}^{i-1}\sin\beta_j)^2 + (\prod_{j=1}^{n-1}\sin\alpha_j-\prod_{j=1}^{n-1}\sin\beta_j)^2 \le \sum_{i=1}^{n-1}(\alpha_i-\beta_i)^2
$$
Proof. Do induction on $n$.
For $n=2$, $$\begin{aligned}\text{LHS}&=(\cos\alpha_1-\cos\beta_1)^2 + (\sin\alpha_1-\sin\beta_1)^2\\
&=2-2\cos(\alpha_1-\beta_1)=4\sin^2(\frac{\alpha_1-\beta_1}2)\le (\alpha_1-\beta_1)^2=\text{RHS.}\end{aligned}$$
For $n+1$, let $\displaystyle{A=\prod_{j=1}^{n-1}\sin\alpha_j}$, $\displaystyle{B=\prod_{j=1}^{n-1}\sin\beta_j}$. Note $|AB|\le1.$
$$\begin{aligned}
\text{LHS } =&\sum_{i=1}^{n}(\cos\alpha_i\prod_{j=1}^{i-1}\sin\alpha_j-\cos\beta_i\prod_{j=1}^{i-1}\sin\beta_j)^2 + (\prod_{j=1}^{n}\sin\alpha_j-\prod_{j=1}^{n}\sin\beta_j)^2 \\
&\quad(\text{Apply the induction hypothesis.})\\
\le&\sum_{i=1}^{n-1}(\alpha_i-\beta_i)^2-(A-B)^2 +(A\cos\alpha_n-B\cos\beta_n)^2+(A\sin\alpha_n-B\sin\beta_n)^2\\
=&\sum_{i=1}^{n-1}(\alpha_i-\beta_i)^2 + (2-2(\cos\alpha_n\cos\beta_n+\sin\alpha_n\sin\beta_n))AB\\
=&\sum_{i=1}^{n-1}(\alpha_i-\beta_i)^2 + (2-2\cos(\alpha_n-\beta_n))AB\\
\le&\sum_{i=1}^{n-1}(\alpha_i-\beta_i)^2 + (2-2\cos(\alpha_n-\beta_n))\\
\le&\sum_{i=1}^{n-1}(\alpha_i-\beta_i)^2 + (\alpha_n-\beta_n)^2\\
=&\text{ RHS} \quad\Box
\end{aligned}$$

Tracing the conditions for the equalities that appear in the proof above, we can verify immediately that $\|\boldsymbol{x}-\boldsymbol{y}\|_2=\|\phi(\boldsymbol{x})-\phi(\boldsymbol{y})\|_2$ iff $\boldsymbol x = \boldsymbol y$.
A: If I reformulate correctly, you express $y \in \mathbb{S}_{n-1}$ as a function $f$ of $\phi = (\phi_1,\ldots,\phi_{n-1}) \in \mathbb{R}^{n-1}$, and you ask whether this function $f$ is $1$-Lipschitz or not. Since $f$ is differentiable, it suffices to check that $||df(\phi)|| \le 1$ is everywhere.
The operator norm $||df(\phi)||$ is the square root of the largest eigenvalue of $df(\phi)^\top \times df(\phi)$.
When $n=3$, the Jacobian matrix is
$$df(\phi) = \left(
\begin{array}{cc}
-\sin \phi_1 & 0 \\
\cos\phi_1 \cos\phi_2 & -\sin\phi_1 \sin\phi_2\\
\cos\phi_1 \sin\phi_2 & \sin\phi_1 \cos\phi_2
\end{array}
\right)$$
One checks that $df(\phi)^\top \times df(\phi)$ is diagonal with diagonal entries $1$ and $\sin^2\phi_1$. My impression is that the same method works in higher dimensions, with heavier formulas.
A: The spherical distance between $x,x'\in\mathcal{S}^{n-1}$ is just $d(x,x')=\arccos(x^Tx')$.
The Euclidean distance is $\|x-x'\|_2=\sqrt{2(1-x^Tx')}$.
So both are just a function of $x^Tx'\in[-1,1]$.
At $x^Tx'=1$, both expressions are zero. Moreover for $t\in(-1,1)$, the derivative of $t\mapsto \arccos(t)$ is smaller than the derivative of $t\mapsto \sqrt{2(1-t)}$. So, $\arccos(t)\geq \sqrt{2(1-t)}$ for all $t\in[-1,1]$.
We conclude that $d(x,x')\geq \|x-x'\|_2$ for all $x,x'\in\mathcal{S}^{n-1}$.
