Prove $\langle x,x \rangle < 0$ or $\langle x,x \rangle > 0$ for all $x \neq 0$ [Added by PLC: This question is a followup to this already answered question.]
Keep the axioms for a real inner product (symmetry, linearity, and homogeneity). 
But make the fourth be
$$\langle x,x \rangle = 0 \text{ if and only if } x = 0.$$
I want to prove that either $\langle x,x \rangle > 0$ or $\langle x,x \rangle < 0$ for all $x \neq 0$. 
Note: $c_1 = \langle x,x \rangle > 0$ and and $c_2 = \langle y,y \rangle < 0$.
Here's the sketch of the proof I want: Assume $\langle x,x \rangle > 0$ for some $x$ and $\langle y,y \rangle < 0$ for some $y$. I'm trying to find a $z \neq 0$ such that $\langle z,z\rangle = 0$, where $z$ is in the space spanned by $\{x,y\}$. By contradiction, we know that $\langle x,x \rangle < 0$ or $\langle x,x \rangle > 0$ for all $x \neq 0$.
Unfortunately, my proof doesn't work that way, and I don't think it proves what I want it to prove.
I say let $\langle z,z \rangle = \langle ax + by, ax + by\rangle$, for $a,b \in \mathbb{R}$. Then
$$\langle z,z \rangle = a^2 \langle x,x \rangle + 2ab \langle x,y \rangle + b^2 \langle y,y \rangle = 0.$$
Let $a = \langle y,y \rangle$ and $b = \langle x,x \rangle$. 
After plugging $a$ and $b$ in, I get: 
\begin{align*}
& \langle y,y \rangle + 2\langle x,y \rangle + \langle x,x \rangle\langle y,y \rangle = 0 \\
\implies& c_2 + 2 \langle x,y \rangle + c_1 c_2 = 0 \\
\implies& 2\langle x,y \rangle = -(c_1 c_2 + c_2).
\end{align*}
Thus,
\begin{align*}
\langle z,z\rangle=
&= c_1 c_2 + 2(-c_1 c_2 - c_2) + c_1 c_2 \\
&= -2c_1c_2 + 2(c_1c_2 c_2) \\
&= -c_1c_2+c_1c_2 +c_2 = 0
\end{align*}
Then $c_2 = 0$
The only thing I can think to do now is to claim a contradiction: we said $c_2 < 0$. But I don't think this proves what we want to prove that either $\langle x,x \rangle < 0$ OR $\langle x,x \rangle > 0$ for all $x \neq 0$.
I think my issue is that I don't know how to choose $a,b$ to make $\langle z,z \rangle = 0$. Someone please offer some help.
 A: Suppose that $\langle x, x \rangle > 0$ and $\langle y, y \rangle < 0$.  I claim that there are $a,b \in \mathbb{R}$ such that $ax+by \neq 0$ and  $\langle ax+by, ax+by \rangle = 0$.  This shows that if 
$q(v) = \langle v, v \rangle$ assumes both positive and negative value, then there is some nonzero $v$ with $q(v) = 0$.
We have $\langle ax+by, ax+by \rangle = a^2 \langle x,x \rangle + 2ab \langle x,y \rangle + b^2 \langle y,y \rangle$.  If we view this as a quadratic equation in $a$, its discriminant is
$\Delta = 4b^2 \langle x,y \rangle^2 - 4 b^2 \langle x,x \rangle \langle y, y \rangle$.  
Because of the assumptions on the sign of $\langle x,x \rangle$ and $\langle y, y \rangle$, $\Delta > 0$ when $b \neq 0$.  So choose your favorite nonzero value of $b$; then the quadratic formula shows that the equation $\langle ax+by,ax+by \rangle = 0$ can be solved for $a$.    
A: We can assume by scaling that $\langle x,x\rangle =1$ and $\langle y,y\rangle=-1$.
The method should work: let $z=x+\lambda y$. Then we have
$$\langle z,z\rangle\ =\ 1+2\lambda\langle x,y\rangle-\lambda^2$$
which has a real root in $\lambda$, as $\langle x,y\rangle$ is considered constant (given).
