Let $V$ be a real vector space, and let $(\cdot,\cdot;\cdot,\cdot) : V^4 \to \mathbb{R}$ be a multilinear form with the following properties:

- $(x,y;z,w) = (y,x;z,w) = (x,y;w,z)$ (symmetry in the first and second pairs)
- $(x,x;z,z) \ge 0$ (positive semidefiniteness in the first and second pairs).

Must such a form satisfy the inequality $$|(x,y;z,w)| \le \sqrt{(x,x;z,z)(y,y;w,w)}?$$

The prototype I have in mind is something like $V = C_c^\infty(\mathbb{R}^n)$, with $$(f,g;h,k) = \int f g \nabla h \cdot \nabla k$$ in which case the inequality follows by using Cauchy-Schwarz twice (first in $\mathbb{R}^n$, and then in $L^2(\mathbb{R}^n)$).

I'd settle for the inequality $$|(x,y;z,w)| \le C({\epsilon}(x,x;z,z) + \epsilon^{-1}(y,y;w,w))$$ which follows from the above by AM-GM (with $C = 1/2$). I'd also settle for the special case $x=w, y=z$ where it reads $|(x,z;x,z)| \le \sqrt{(x,x;z,z)(z,z;x,x)}$.

~~Simply using Cauchy-Schwarz in each pair gives the inequality
$$|(x,y;z,w)| \le [(x,x;z,z)(x,x;w,w)(y,y;z,z)(y,y;w,w)]^{1/4}$$ which has cross terms that I don't want.~~ Edit: Of course, as Willie Wong points out and zeb's counterexample confirms, this doesn't work.

Thanks!