Let $L$ be a number field, and let $\alpha\in L$. We say that $\alpha\in L$ is totally positive if $\sigma(\alpha)$ is a positive real number for all $\mathbb{Q}$-embeddings $\sigma:L\hookrightarrow \mathbb{C}$.
The following theorem is known:
Thm. Let $L$ be an number field. Then $\alpha\in L$ is totally positive if and only if $\alpha$ is a sum of squares in $L$.
This theorem follows from the following two facts.
If $L$ is a formally real field, then the intersection of all orderings of $L$ is precisely the set of sums of squares (easy)
If $L$ is a number field, the orderings of $L$ are the $\sigma^{-1}(\mathbb{R}_{>0})$, where $\sigma$ runs through the set of embeddings of $L$. (not that easy)
The second fact is not that easy, and rather technical. Its is using the notion of real closure, extensions of embeddings for quadratic field extensions... so:
Question. Is it possible to find a direct proof of theorem (direct= not using the theory of ordered fields) ?
Note that in the statement of the theorem, we may reduce to $L=\mathbb{Q}(\alpha)$, since any embedding of $\mathbb{Q}(\alpha)$ is obtained by restriction of an embedding of $L$.
In this case, one may reformulate the theorem as follows: let $f$ be an irreducible polynomial of $\mathbb{Q}[X]$. Assume that all real roots of $f$ are positive. Then any real root $\alpha$ is a sum of squares in $\mathbb{Q}(\alpha)$.
I may answer to the question in a special case: the minimal polyomial of $\alpha$ has only real roots and has even degree.
Assume that $f=\prod_{i=1}^{2n}(X-\alpha_i)$ where $\alpha_i>0$. Separating even and odd powers of $X$, we get $f=X^{2n}+P(X^2)-Q(X^2)X$, where $P,Q$ are polynomials with positive rational coefficients of degree $< n$. Now if $\alpha$ is a root of $f$, $\alpha=\dfrac{P(\alpha^2)Q(\alpha^2)}{Q(\alpha^2)^2}$. But positive rational numbers are sum of squares. Since products of sum of squares are sum of squares, we have the desired conclusion.
I have no idea on how to deal with the general case. I would be happy if someone could come up with any idea.