Let $V$ be a real vector space. Let $\langle \cdot , ..., \cdot \rangle : V \times \cdots \times V \rightarrow \mathbb{R}$ be a symmetric, positive definitve multilinear map in $p$ variables.**
Question: Define a map $|| \cdot || : V \rightarrow \mathbb{R}$ sending $v$ to $\sqrt[p]{|\langle v, ..., v \rangle|}$. Is $|| \cdot ||$ a norm?
We have have everything except possibly the triangle inequality, so let's try to show that $||v + w || \leq ||v|| + ||w||$.
It seems like we want a generalized Cauchy-Shwartz inequality:
$$\langle v_1, ..., v_n \rangle \leq || v_1 || \cdot \cdots \cdot ||v_n||$$
If that is satisfied, then
$$|| v + w || = \sqrt[p]{\left| \sum_{t_1, ..., t_n \in \{ v, w \} } \langle t_1, ..., t_n \rangle \right| } $$ $$\leq \sqrt[p]{ \sum_{t_1, ..., t_n \in \{ v, w \} } ||t_1 || \cdots ||t_n|| } = \sqrt[p]{(||v|| + ||w||)^p} = ||v|| + ||w||$$
If $p = 2$, then this is true.
** Note: by symmetric I mean that $\langle v_1, ..., v_p \rangle = \langle v_{\sigma(1)}, ..., v_{\sigma(p)} \rangle$ for each permutation $\sigma \in S_p$. By nondegenerate I mean $\langle v, v, ..., v \rangle = 0 \implies v = 0$.