Cauchy-Schwarz inequality for inner products
If $V$ is a real vector space and $f: V\times V\to \mathbb{R}$ is a symmetric bilinear positive map, then we have the Cauchy-Schwarz inequality $$f(v,w)^2\le f(v,v)f(w,w)\text{ for all }v,w\in V,$$ which is proved for example by examining the discriminant of the quadratic function $$f(Xv+w,Xv+w)=f(v,v)X^2+2f(v,w)X+f(w,w).$$
A generalization ?
Now let $V$ is a $\mathbb{Z}$-module and $f: V\times V\to \mathbb{R}$ a symmetric bilinear positive function such that $f(nv,mw)=nmf(v,w)$ for $v,w\in V$ and $n,m\in\mathbb{Z}$.
The question is: Do we still have a Cauchy-Schwarz inequality on $f$ ?
The idea of the proof above can be used to prove that $f(v,w)^2\le f(v,v)(f(v,v)/4+f(w,w))$, but we can't seem to do better with this idea since $\mathbb{Z}$ itself is not a field.
The same proof gives $4 \times$ (Cauchy Schwarz inequality). Because the values of $f$ considered in the inequality are real numbers, not elements of the $Z$-module, division by $4$ is possible and the factor of $4$ can be removed.