Let $E$ be an Euclidean space with norm $\left\lVert \cdot \right\rVert$. (In this post, a normed space is called "Euclidean" if its norm comes from a scalar product, or equivalently, if the parallelogram law is satisfied).
If $B(u,v)$ is a bilinear form on $E$ satisfying $$ \left\lvert B(u,u) \right\rvert \leq c \left\lVert u \right\rVert^2 \label{eq1}\tag{1} $$ for all $u \in E$, then $$ \left\lvert B(u,v) \right\rvert \leq c \left\lVert u \right\rVert\,\left\lVert v \right\rVert \label{eq2}\tag{2} $$ for all $u,v \in E$. (A proof: https://mathoverflow.net/questions/207998/norm-of-n-linear-symmetric-forms/208025#208025?newreg=8d0314c729f840789c615a033e5fad50)
We wonder whether the reverse holds: if (\ref{eq1}) $\implies$ (\ref{eq2}) for all bilinear forms $B$, does it means that the space is Euclidean? We have not been able to find any counter-example.
Regards.
Here are some ideas. Maybe somebody else will be able to develop them into an actual answer.
Assume that vectors $v,w$ and subspace $S$ of codimension $2$ are such that the sum $Span(v,w)\oplus S$ is the whole vector space and that $\|av+bw+s\|\geq \max(|a|\|v\|, |b|\|w\|)$ for all $a,b\in\mathbb{R}$ and $s\in S$. Assume also that $\|v+w\|^2 + \|v-w\|^2 < 2(\|v\|^2 + \|w\|^2)$. Then the form $B(a_1v+b_1w+s_1, a_2v+b_2w+s_2) = a_1a_2\|v\|^2 - b_1b_2\|w\|^2$ is a counterexample: $|B(u,u)|\leq \|u\|^2$ for all $u$, but $B(v+w, v-w) = \|v\|^2 + \|w\|^2 > \|v+w\|\|v-w\|$.
As $v$ and $w$ for $\mathbb{R}^2$ and $\|\cdot\|_p$, one can take $(1,0)$ and $(0,1)$ is $p>2$, and $(1,1)$ and $(1,-1)$ if $p<2$. I don't know if it's possible to find such $v, w$ and $S$ for any non-Euclidean normed space.