Necessary conditions for Cauchy-Schwarz like inequality I did a quick search and did not find much.  I was wondering what conditions on $f$ and a space $X$ are necessary in order to have
$$f(x,x)f(y,y)\ge f(x,y)^2\quad \forall\ x,y\in X.$$ Clearly if $f$ is an inner-product and $X$ is a Hilbert space it is true, but from my knowledge, this is not necessary.  What clearly is necessary is that if $f(x,y)\ne0$ then $f(x,x)\ne0$ and $f(y,y)\ne0$.  Does anyone know any other necessary conditions, or looser sufficient conditions?  Thanks.
I was asked for more context.  Though this is probably too narrow, I was looking at a sort of composition of C-S.  That is, given vectors $x_1,x_2,x_3,x_4$ in a Hilbert Space (Euclidean if you wish), I want to show something like
$$(\lVert x_1\rVert^2\lVert x_2\rVert^2-\langle x_1,x_2\rangle^2)(\lVert x_3\rVert^2\lVert x_4\rVert^2-\langle x_3,x_4\rangle^2)-(\langle x_1,x_3\rangle\langle x_2,x_4\rangle-\langle x_1,x_4\rangle\langle x_2,x_3\rangle)^2\ge0$$
In my current context, $x_1=x_3$, but I have a feeling that is not necessary, though I am not 100% confident this is a true statement.  If this is not enough details, then $x_1=\{\alpha^k\}_{k=0}^n$, $x_2=\{k\alpha^{k-1}\}_{k=0}^n$ and $x_4=\{\beta^k\}_{k=0}^n$ for some $n$ and $\alpha,\beta\in\mathbb{R}$.
 A: As the OP explains, he is interested with what he calls a composition of Cauchy-Schwarz inequality, and this is what I propose to prove.
Lemma. Suppose that $M=\left[\matrix{A&B\cr B^*& C}\right]$ is a block $2n\times 2n$ matrix which is positive then
$$ |\det B|^2\le \det A \det C$$
Proof. We will suppose that $M$ is Positive definite, the general case follows by a simple limiting argument.
Let us consider the Choleski decomposition of $M$ :
$$\left[\matrix{A&B\cr B^*& C}\right]=
\left[\matrix{S^*&0\cr T^*& R^*}\right] \left[\matrix{S&T\cr 0& R}\right]$$
where $S$ and $R$ are upper triangular. Now, we have $$A=S^*S,\quad B=S^*T,\quad C=T^*T+R^*R$$
So
$$\det (B^*B)=\det (S^*S)\det(T^*T)=\det A\det(T^*T)$$
Now since $C-T^*T=R^*R\ge0$ we conclude that $\det C\ge \det( T^*T)$, and consequently $|\det B|^2=\det (B^*B)\le \det A\det C$.$\square$
Now if we apply this to the Gram matrix 
$\def\aa#1#2{\langle x_#1,x_#2\rangle}$
$$M=\left[\matrix{\Vert x_1\Vert^2& \aa12&\aa13&\aa14\cr
\aa21&\Vert x_2\Vert^2&\aa23&\aa24\cr
\aa31&\aa32&\Vert x_3\Vert^2 &\aa34\cr
\aa41&\aa42&\aa43&\Vert x_4\Vert^2
}\right].$$
We obtain the desired inequality.$\square$
