Suppose \(\mathbb{F} = \mathbb{R}\), and \(V \neq \{0\}\). Replace the positivity condition with the condition that \(\langle v,v \rangle > 0\) for some \(v \in V\). Show that this change in definition does not change the set of functions from \(V \times V\) to \(\mathbb{R}\) that are inner products on \(V\).
We hope to show that \(\langle v,v \rangle >0\) for some \(v \in V\) implies that \(\langle v,v \rangle \geq 0\) for all \(v \in V\) in real vector spaces.
Take some \(v_0 \in V\) such that \(\langle v_0,v_0 \rangle >0\). Now, WLOG let \(v \in V\) and \(v = v_0+w\). So:
\begin{align} 0 &< \langle v_0,v_0 \rangle \\ &= \langle v-w, v-w \rangle \\ &= \langle v,v \rangle + \langle w,w \rangle - 2\langle v,w \rangle \end{align}
Now, the last step is possible because symmetry becomes conjugate symmetry in reals.
We now have that:
\begin{equation} 2 \langle v,w \rangle - \langle w,w \rangle < \langle v,v \rangle \end{equation}