Does Cauchy-Schwarz Inequality depend on positive definiteness? Let $V$ be a vector space over $\mathbb{R}$. Suppose we have a product $\langle \cdot,\cdot\rangle:V^2\to \mathbb{R}$ that satisfies all the inner product axioms except the second part of positive-definiteness: $$\langle x,x\rangle=0\iff x=0\tag{1}$$
So far, every proof I've seen that the Cauchy-Schwarz Inequality holds for all inner product spaces uses $(1)$. But does a proof of the Cauchy-Schwarz Inequality necessarily depend on $(1)$? Specifically, I'm looking for one of the following:


*

*A proof that the Cauchy-Schwarz Inequality holds for all "inner product spaces" where $(1)$ does not necessarily hold.

*A counterexample of a product $\langle \cdot,\cdot\rangle$ that follows symmetry, linearity in the first parameter, and $\langle u,u\rangle\ge 0$ for all $u\in V$, but where $\lvert \langle u,v\rangle\rvert\le \lvert\lvert u\rvert\rvert\ \lvert\lvert v\rvert\rvert$ does not always hold.


I suspect that there is a counterexample, but it's hard for me to come up with one.
 A: Yes, Cauchy-Schwartz definitely depends on
 definiteness. It fails for indefinite forms: one can have
$\left<u,u\right>=0$ but $\left<u,v\right>>0$. For instance consider
$\left<(a,b),(c,d)\right>=ac-bd$ and $u=(1,1)$ and $v=(1,-1)$.
A: If $\|a\| = \|b\| = 0$, then
\begin{align*}
0 & \leqslant \|a + b\|^2 =
\|a\|^2 + \|b\|^2 + 2\langle a, b \rangle =
+2\langle a, b \rangle, \\
0 & \leqslant \|a - b\|^2 =
\|a\|^2 + \|b\|^2 - 2\langle a, b \rangle = -2\langle a, b \rangle,
\end{align*}
therefore $\lvert\langle a, b \rangle\rvert = 0 \leqslant \|a\|\|b\|$.
Suppose, on the other hand, that $\|b\| > 0$. Then a fairly standard argument applies. Define $\lambda = \langle a, b \rangle/\|b\|^2$. Then
$\langle a - \lambda b, b \rangle = \langle a, b \rangle - \lambda \|b\|^2 = 0$, therefore
\begin{align*}
0 & \leqslant \langle a - \lambda b, a - \lambda b \rangle  \\
& = \langle a, a - \lambda b \rangle - \lambda\langle b, a - \lambda b \rangle \\
& = \langle a, a - \lambda b \rangle - \lambda\langle a - \lambda b, b \rangle \\
& = \langle a, a - \lambda b \rangle \\
& = \|a\|^2 - \lambda \langle a, b \rangle,
\end{align*}
therefore
$$
\langle a, b \rangle^2 = \lambda\|b\|^2\langle a, b \rangle \leqslant \|a\|^2\|b\|^2,
$$
therefore
$$
\lvert\langle a, b \rangle\rvert \leqslant \|a\|\|b\|.
$$
The argument is similar when $\|a\| > 0$. So the Cauchy-Schwarz inequality holds in all cases.
Addendum
It appears (see my series of shame-faced comments below for details)
that this argument is merely an obfuscation of what is surely the
most "standard" of all proofs of the Cauchy-Schwarz inequality. It is
the one that is essentially due to Schwarz himself, and he had good
reasons for using it, quite probably including the fact that it makes
no use of the postulate that $\langle x, x \rangle = 0 \implies x = 0$!
In a modern abstract formulation, it goes as follows (assuming, of course,
that I haven't messed it up again). For all real $\lambda$, we have
$\|u\|^2 - 2\lambda\langle u, v \rangle + \lambda^2\|v\|^2 =
\|u - \lambda v\|^2 \geqslant  0$.
Therefore, the discriminant of this quadratic function of $\lambda$
must be $\leqslant 0$. That is, $\langle u, v \rangle^2 \leqslant \|u\|^2\|v\|^2$; equivalently, $\lvert\langle u, v \rangle\rvert \leqslant \|u\|\|v\|$.
