Let $\mathbb{S}^{n}$ be the canonical simplex of $\mathbb{R}^{n}$ and let $u = (1/n,\dotsc,1/n)$. Is it true that
$$\lVert x - u \rVert_p \leq \lVert y - u \rVert_p$$
implies that
$$\lVert x\rVert_p \leq \lVert y\rVert_p$$
for all $x,y \in \mathbb{S}^{n}$? It seems intuitive to me that this proposition is indeed true because the $p$-norm is minimized by $u$ in the $n$-simplex. However, I can't find a formal proof for that. Any ideas?
The answer is yes if $p=2$ (by the Pythagoras theorem: $\|x\|_2^2=\|x-u\|_2^2+\|u\|_2^2$, because $x-u\perp u$) and, obviously, if $p=1$ or $n\le2$.
Otherwise, the answer is no, because then the value of $\|x-u\|_p$ does not determine the value of $\|x\|_p$.
Details on the last sentence: Suppose that $n\ge3$ and $p\in(0,\infty)\setminus\{1,2\}$. For small $t,s>0$, let
$$x(t):=u+t(1,-1,0,\dots,0),\quad y(s):=u+s(1,-1/2,-1/2,0,\dots,0).$$
Then $x(t)$ and $y(s)$ are in the simplex and
$$\|x(t)-u\|_p=\|y(s)-u\|_p$$
if
$$t=c_p s,\quad\text{where}\quad c_p:=(2^{-p}+1/2)^{1/p},$$
which latter will be assumed henceforth.
Then (for $s\downarrow0$),
$$\|x(t)\|_p^p-n^{1-p}\sim n^{2-p}p(p-1)t^2=n^{2-p}p(p-1)c_p^2 s^2,$$
$$\|y(s)\|_p^p-n^{1-p}\sim \tfrac34\,n^{2-p}p(p-1)s^2.$$
So, if your conjecture were true, we would have $\|x(t)\|_p=\|y(s)\|_p$ and hence $c_p^2 =\tfrac34$, which would imply $p\in\{2,4\}$. But $p=2$ was excluded. It remains to exclude $p=4$.
But, if $p=4$, then $\|x(t)\|_p^p-\|y(s)\|_p^p\sim-3s^3/n$, so that $\|x(t)\|_p\ne\|y(s)\|_p$, the final contradiction.
Further detail, on why $c_p^2 =\tfrac34$ implies $p\in\{2,4\}$. Rewrite $c_p^2 =\tfrac34$ as $g(p):=2^{p-1}-3^{p/2}+1=0$. Note that $2\times3^{-p/2}g'(p)=(4/3)^{p/2}\ln2-\ln3$ is increasing in real $p>0$ from $\ln2-\ln3<0$ to $\infty$. So, for some real $p_*>0$, the function $g$ is decreasing on $(0,p_*]$ and increasing on $[p_*,\infty)$. So, $g$ has at most two positive roots. Also, $2$ and $4$ are roots of $g$, and hence the only positive roots of $g$.
