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?

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?

Let $\mathbb{S}^{N}$ be the canonical simplex of $\mathbb{R}^{N}$ and let $u = (1/N,...,1/N)$. Is it true that

$|| x - u ||_p \leq || y - u ||_p$

implies that

$||x||_p \leq ||y||_p$

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?

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?