A Jordan Algebra $(V,*)$ is an algebra with a commutative (not associative) multiplication operator $*$ that satisfies the Jordan identity: $(x*y)*(x*x) = x*(y*(x*x))$. The positive cone consists of the squares $x^2=x*x$. We call a Jordan Algebra Euclidean if it has a symmetric inner product to which the positive cone is self-dual.
We can represent the multiplication using the multiplication operator $L(x):V\rightarrow V$, $L(x)y = x*y$.
The quadratic representation of the Jordan Algebra is then given by $Q(x)=2L(x)^2 - L(x^2)$.
The canonical example of a Jordan Algebra is that of a matrix algebra of self-adjoint matrices $M_n(\mathbb{C})^{sa}$ equipped with the product $$x*y = \frac{1}{2}(xy+yx)$$
With this multiplication the quadratic form is $Q(x)y = xyx$, so in particular, when $x,y\geq 0$ we have $Q(x)y\geq 0$.
My question is: Is it true in an arbitrary Euclidean Jordan Algebra that when $x,y\geq 0$ we have $Q(x)y\geq 0$?
Okay, so apparently I didn't look hard enough for an answer. The Wikipedia page on symmetric cones has a section on the quadratic representation where they give a proof of this statement. I would welcome a more direct or simple proof however.
Wikipedia: Symmetric cone