History of "no positive definite ternary integral quadratic form is universal"? 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!
 A: Hi, Pete. There are a few observations related to this, not widely known although basic, and that includes your colleague. First, Conway gives a quick proof on page 142 of The Sensual Quadratic Form, including over the rationals. 
Next, also Conway, the form (five variables) that he and Schneeberger found that represents all the numbers from 1 to 289, fails to represent 290, then represents 291 and on forever, he initially called Methusaleh. It is just a binary added to a ternary that represents the numbers from 1 to 28 consecutive, discriminant 29. However, for ternaries that is not the record. The form he called Little Methusaleh, discriminant 31, represents 1 to 30 consecutive. The theorem is in this material, as the conditions for a positive ternary to represent, say, 1,2,3,5, places strong restrictions on a partly reduced form. Kap wrote this sort of argument up several times, including a repeat in the unpublished 1996 Classification. It is quite easy. OK, Little Methusaleh and your result over the integers are proved on page 81 of  The Sensual Quadratic Form
Finally, a positive form is anisotropic at the "prime" infinity. In Cassels Rational Quadratic Forms he shows global relations on the Hilbert Norm Residue symbol that show that any ternary is anisotropic at an even number of primes. So a positive ternary is anisotropic at an odd number of finite primes. Taken with the observation above that at least one number below 31 is missed, and a positive ternary fails to integrally represent an infinite number of positive integers.
I will look up some of my tables and fill things in. Note that some of this is discussed in an early article by William Duke, 1997 Notices, but he mistyped the form with discriminant 29.
Let's see, Conway and Schneeberger probably had an acceptable proof of the 15 Theorem scattered about, but it never got put together. Bhargava was looking for diversions from his own dissertation, Conway mentioned this in passing. Bhargava showed the fundamental result that one of these forms must have a regular ternary as a sub-form, thus the project became a careful inspection of my paper with Kap on all possible regular ternaries. Also, correspondence between Kap and Bhargava first revealed some important errors in Magma relating to calculating the spinor genus, and hilarity ensued.
EDIT: thinking about the history question, it is quite possible that this result was never written down as a separate proposition, by Gauss, Legendre, etc. The reason I suggest this is the great weight placed on positive ternary forms missing certain "progressions," in the language of Jones, Dickson, other early books. So, in Jones, chapter 8, we read "Thus there will be a finite number of arithmetical progressions of this type" of numbers not represented by any form in the genus under consideration. Not much motivation for proving that a form misses at least one number if you are going to quickly show that it misses an entire arithmetic progression.
EDIT TOOO: note that Conway replaces the prime usually called $\infty$ by the prime $-1.$

No definite ternary form is universal
However, a simple argument shows that
  any definite ternary form must fail to
  represent infinitely many integers,
  even over the rationals. For if a
  ternary form $f$ of determinant $d$
  represents anything in the $p$-adic
  squareclass of $-d$ over $\mathbf
> Q_p,$ then it must be $p$-adically
  equivalent to $[ -d,a,b]$ where the
  "quotient form" $[a,b]$ has
  determinant $-1,$ and so $p$-adically,
  $f$ must be the isotropic form $[
> -d,1,-1].$
But a positive definite form fails to
  represent $-1,$ and so it is not 
  $p$-adically  isotropic for $p=-1.$ By
  the global relation, there must be
  another $p$ for which it is not 
  $p$-adically  isotropic, and so it
  also fails to represent all numbers in
  the $p$-adic square-class of $-d$ for
  this $p$ too!
The Three Squares Theorem illustrates
  this nicely--the form $[ 1,1,1]$ fails
  to represent $-1$ both $-1$-adically
  and $2$-adically. In the Third
  Lecture, we showed that The Little
  Methusaleh Form  $$  x^2 + 2 y^2 + y z
> + 4 z^2     $$ fails to represent 31. We now see that since it fails to
  represent the $-1$-adic class of its
  determinant $-31/4$ (i.e., the
  negative numbers), it must also fail
  to represent the infinitely many
  positive integers in the $31$-adic
  squareclass of  $-31/4.$

A: Greg Doyle and Kenneth Williams have more recently (2017) given a completely elementary proof: "A POSITIVE-DEFINITE TERNARY QUADRATIC FORM DOES
NOT REPRESENT ALL POSITIVE INTEGERS"
