Let $V$ be a vector space over $\mathbb R$ and $g:V\times V\rightarrow \mathbb R$ be a bilinear form on $V$. The usual statement for the generalized Cauchy inequality usually goes like this:
Let $g$ be such that
- $g(v,v)\ge0$ for every $v\in V$.
- $g(v,w)=g(w,v)$ for every $v,w \in V$.
Then $$g(v,w)^2 \le g(v,v)g(w,w)$$
The opposite implication is false; for instance, if we consider $\mathbb R^2$ equipped with the form $g$ having matrix $$A=\begin{pmatrix}1&0\\0&-1 \end{pmatrix}$$ and two vectors $v=(v_1,v_2)^T,w=(w_1,w_2)^T$, then
$$g(v,w)^2=(v_1w_1-v_2w_2)^2=v_1^2w_1^2+v_2^2w_2^2-2v_1w_1v_2w_2 $$ $$g(v,v)g(w,w)=(v_1^2+v_2^2)(w_1^2+w_2^2) =v_1^2w_1^2+v_1^2w_2^2+v_2^2w_1^2+v_2^2w_2^2$$ So \begin{array}{l l} & g(v,w)^2 \le g(v,v)g(w,w) \\ \iff & v_1^2w_2^2+v_2^2w_1^2+2v_1w_1v_2w_2\ge 0 \\ \iff &(v_1w_2+v_2w_1)^2\ge 0 \end{array} which is clearly true. However, $g$ does not satisfy the first hypothesis.
Is there any easy way to characterize the bilinear forms $g$ such that the Cauchy-Schwarz inequality is true in the space $(V,g)$?