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This is surely trivial, but I've never seen a triangle explicitly defined in this way before. Let $a,b,c>0$. Is it true that:

$$a,b,c\text{ form a triangle}\iff -1\leq\frac{b^2+c^2-a^2}{2bc}\leq 1.$$ My attempt: ($\implies$) Suppose $a,b,c$ form a triangle, then by the law of cosines, $$\cos \vartheta=\frac{b^2+c^2-a^2}{2bc},$$ and $-1\leq \cos \vartheta\leq 1$.

($\impliedby$) Suppose $$\frac{b^2+c^2-a^2}{2bc}=\lambda,$$ with $-1\leq\lambda\leq 1$, then there exists a $\vartheta$ such that $\cos\vartheta=\lambda$ and so $$\frac{b^2+c^2-a^2}{2bc}=\cos\vartheta.$$ Hence, by the law of cosines $a,b,c$ form a triangle.

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    Side-angle-side defines a triangle (well one way). The law of cosines give the trigonometry to find the remaining side when you have SAS. – 2017-01-04
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    @Doug M thanks. I'm interested to know when three lengths form a triangle, but I want to do this via the law of cosines if possible, not via e.g. $a+b>c$, $a+c>b$, $b+c>a$. In particular I'm interested in the $\impliedby$ part. I know the $\implies$ part is correct. – 2017-01-04
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    I think what Doug said explains the $\impliedby$ direction (allowing for degenerate triangles with angles $\pi$, $0$, and $0$ if the fraction is $\pm1$). If $-1\leq\frac{b^2+c^2-a^2}{2bc}\leq 1$, let $\theta=\arccos\frac{b^2+c^2-a^2}{2bc}$ and construct the triangle using sides $b$ and $c$ around angle $\theta$. There is such a triangle for any $b$, $c$, and nonnegative $\theta$ less than or equal to $\pi$ – 2017-01-04

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Let $a,b,c>0$ and

$$ -1\leq\frac{b^2+c^2-a^2}{2bc}\leq 1. \tag1$$

Then your proof via the Law of Cosines requires the existence of $\vartheta$ such that $0 \leq \vartheta \leq \pi$ and $\cos\vartheta = \frac{b^2+c^2-a^2}{2bc}.$ Such a value $\vartheta$ always exists under the conditions given in Inequality $1$. Then you need merely construct a triangle with sides $b$ and $c$ adjacent to angle $\vartheta$, and the Law of Cosines then implies that the opposite side will have length $a$. Hence a triangle with sides $a,b,c$ exists.

This part of the answer (which I think is what you wanted) was already given in comments, so I'm not taking credit for it. Someone else can post a non-community-wiki version if desired.

Alternatively, starting with $a,b,c>0$ and Inequality $1$, it is a fact that $-2bc < 0$, so multiplying everything by $-2bc$ reverses the directions of both inequalities:

$$ 2bc \geq a^2 - b^2 - c^2 \geq -2bc.$$

Add $b^2 + c^2$:

$$ (b+c)^2 = b^2 + 2bc + c^2 \geq a^2 \geq b^2 - 2bc + c^2 = (b - c)^2.$$

The inequalities are preserved by taking the positive square root, so

$$ b+c \geq a \geq \lvert b - c \rvert,$$

observing that $b+c>0$ and $a>0$ but the sign of $b-c$ has not been determined. The left-hand inequality immediately gives us one of the necessary triangle inequalities, $a \leq b + c$. The right-hand inequality gives us two inequalities, \begin{align} b - c &\leq a,\\ c - b &\leq a,\\ \end{align} from which we get

\begin{align} b &\leq a + c,\\ c &\leq a + b,\\ \end{align} which are the other two triangle inequalities.