Suppose $(E,\la.,.\ra)$ is euclidean and $z\in E$ a unit vector. Then
$$
g(x,y)\colon=\la x,y\ra-2\la x,z\ra\la y,z\ra
$$
is a Lorentz product on $E$. 2. Conversely if $(E,\la.,.\ra)$ is a Lorentz space and $z\in E$ a time-like unit vector, then
$$
g(x,y)\colon=\la x,y\ra+2\la x,z\ra\la y,z\ra
$$
is a euclidean product on $E$ and for all $x,y\in E$ we have:
$$
|\la x,y\ra|\leq3|x||y|
\quad\mbox{where}\quad
|x|^2\colon=g(x,x)~.
$$
2. Put $x=-\la x,z\ra z+u$ for some $u\in z^\perp$, then
$$
g(x,x)
=\la x,x\ra+2\la x,z\ra^2
=-\la x,z\ra^2+\Vert u\Vert^2+2\la x,z\ra^2
=\la x,z\ra^2+\Vert u\Vert^2
$$
and this equals $0$ iff $u=0$ and $\la x,z\ra=0$, i.e. $x=0$ and thus $g$ is a euclidean product. Since $g(x,z)=-\la x,z\ra$ and $|z|^2=g(z,z)=1$ we get by Cauchy-Schwarz:
\begin{eqnarray*}
|\la x,y\ra|
&\leq&|g(x,y)|+2|\la x,z\ra\la y,z\ra|\\
&=&|g(x,y)|+2|g(x,z)g(y,z)|
\leq|x||y|+2|x||y||z|^2
\leq3|x||y|~.
\end{eqnarray*}