Let $\| \cdot \|$ be a norm on $\mathbb{R}^n$. We call it axes-symmetric if $\|x\|$ does not depend on the order of the components of $x$. Equivalently if $\|x\| = \|P \cdot x \|$ for any permutation matrix.
Question: If a norm is axes-symmetric, is it invariant under signed permutation matrices? (i.e $\|(x_1,...,x_n)\| = \|(\pm x_1,...,\pm x_n) \|$ ).
Take $$\|(x_1,x_2)\|^2 = x_1^2+x_1x_2+x_2^2.$$ This relation defines a norm on $\Bbb R^2$.
Clearly, $$\|(x_1,x_2)\|=\|(x_2,x_1)\|\ne \|(x_1,-x_2)\|.$$
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Write this norm in terms of a matrix $A=\frac 12 \begin{pmatrix}2&1\\1&2\end{pmatrix}$ and scalar product:
$$\|\vec x\|^2 = \vec x^TA\vec x = (A\vec x,\vec x).$$ The matrix $A$ is positive definite (symmetric with eigenvalues $\frac 12$ and $\frac 32$), hence the map $(\vec x,\vec y)\to (A\vec x,\vec y)$ defines an inner product. This ensures us that the map $\vec x\to \sqrt{(A\vec x,\vec x)}$ is a norm. This norm is called sometimes the energy norm given by the positive definite matrix $A$.
In general case $A=A^T>0$, proving the triangle inequality for the corresponding energy norm is equivalent to proving that $\vec x^TA\vec y\le \sqrt {\vec x^TA\vec x}\sqrt {\vec y^TA\vec y}$, which can be obtained by applying the Cauchy-Schwarz inequality to scalar product of vectors $\sqrt A \vec x$ and $\sqrt A\vec y$. The matrix $\sqrt A$ is defined as a unique symmetric positive definite matrix $B$ such that $B^2=A$.