I am currently leading a graduate student research group on geometry of numbers (henceforth GoN) and its Diophantine applications, especially to quadratic forms. In an early lecture I gave a bit of a crash course on quadratic forms over $\mathbb{Z}$, $\mathbb{Q}$ and $\mathbb{R}$. In particular, since one of the two most classical GoN proofs is of the Four Squares Theorem, an early goal of our group is to look into the prospect of proving the universality of other integral quadratic forms by GoN methods. (If you care, as of now we have used GoN to prove universality of exactly one other quaternary form.)
Anyway, I was trying to give my students the "facts of life" about positive definite integral quadratic forms which are "positive universal", i.e., integrally represent all positive integers. The big news here is the 15 Theorem of Conway-Schneeberger(-Bhargava?) and the 290 Theorem of Bhargava-Hanke. When I mentioned that these theorems lead to the enumeration of the finite list of universal quaternary forms, a quick student pointed out that there must not be any positive definite universal ternary integral quadratic forms (a result which I had been planning to mention later). So I decided to assign someone to present a proof of this result on ternary forms.
Surprisingly to me, although the statement of the result is quite common, a cursory search of the literature did not reveal a "standard proof". Eventually I remembered / revealed the following proof that I somehow had in mind.
Let $q(x,y,z)$ be a positive-definite, positive universal ternary integral quadratic form.
Step 0: It is easy to see that as a quadratic form over $\mathbb{Q}$, $q$ is still positive-definite and positive-universal, i.e., it $\mathbb{Q}$-represents every positive rational number. We will in fact show that there are no such ternary forms over $\mathbb{Q}$ (a stronger result).
Step 1: In particular $q$ is anisotropic over $\mathbb{R}$, hence by the reciprocity law for Hilbert symbols it is also anisotropic over $\mathbb{Q}_p$ for some $p$. A weak approximation argument shows that since $q$ is positive universal over $\mathbb{Q}$, it is universal over $\mathbb{Q}_p$: i.e., it $\mathbb{Q}_p$-represents every nonzero element of $\mathbb{Q}_p$.
Step 2: I claim that for any field $K$ of characteristic not $2$, there is no anisotropic universal ternary form $q$ over $K$. Indeed, if there is such a form, then for any $\alpha \in K^{\times}$ the scalar multiple form $\alpha \cdot q$ is also anisotropic and universal. By choosing $\alpha$ to be the discriminant of $q$, we may assume that $q$ has discriminant $1$ (Here, as usual in the algebraic theory of quadratic forms, the discriminant of $q$ is a square class, i.e., an element of $K/K^{\times 2}$.) But now it is a basic fact that if $q$ is an anisotropic ternary quadratic form of discriminant $1$, then $q' = q \oplus \langle 1 \rangle = q(x,y,z) + w^2$ is also anisotropic. But since $q$ is universal, it represents $-1$ and therefore $q'$ is visibly isotropic, a contradiction.
Remark 1: I think of the above fact in terms of the ternary norm form and the quaternary norm form associated to a quaternion algebra over $K$. See e.g. Theorem 95 in these notes on non-commutative algebra for a proof. But in fact if you look at the proof you see that quaternion algebras are not mentioned, so it really is an elementary but not completely trivial fact from the algebraic theory of quadratic forms.
Remark 2: Note that Step 2 shows in particular that there is no field $K$ with u-invariant $3$. At the moment, the above argument is my favorite proof of this.
Remark 3: An equivalent statement of the result of Step 2 is that $q(x,y,z) = ax^2 + by^2 + cz^2$ represents $-abc$ iff $q$ is isotropic. This is Exercise 6 in Chapter III of Lam's book on quadratic forms over fields. I am slightly surprised that he does not make a bigger deal of it.
[Remark 4: The argument given by Conway in Will Jagy's answer gives a more elementary solution to Lam's exercise in the above remark than the one I referred to. Namely, if $q$ represents $-\operatorname{disc}(q)$, then $q \cong [-\operatorname{disc}(q),a,b]$, and checking discriminants gives $\operatorname{disc}([a,b]) = -1$. But a binary form has discriminant $-1$ iff it is isomorphic to the hyperbolic plane, so this implies that $q$ is isotropic.]
Finally, my question: what is the "classical proof" of the above theorem about positive-definite ternary integral quadratic forms? How far back does this result go? I would not be at all surprised to learn that it was known to Lagrange / Legendre / Gauss, but presumably they would not have proved it as I did above!
