= Solution
The single bonds give equal simple-root lengths and inner products, after normalization,
$$
(\alpha,\alpha)=(\beta,\beta)=(\gamma,\gamma)=2,\quad
(\alpha,\beta)=(\beta,\gamma)=-1,\quad(\alpha,\gamma)=0.
$$
The <positive roots> of this <A3 root system> are
$$
\boxed{\alpha,\ \beta,\ \gamma,\ \alpha+\beta,\ \beta+\gamma,\ \alpha+\beta+\gamma.}
$$
The first two sums are <roots> by the qualified obtuse-root lemma. The last is a <root> because $(\alpha+\beta,\gamma)=-1$. All have nonnegative simple-root coordinates, hence are positive.
To prove completeness, first note that every <root> is Weyl-conjugate to a <simple root>. For a nonsimple <positive root> $\delta=\sum c_i\alpha_i$, some $(\delta,\alpha_i)>0$, since $(\delta,\delta)=\sum c_i(\delta,\alpha_i)>0$. Reflection subtracts a positive integer multiple of $\alpha_i$, decreasing its integer height. The reflected <root> stays positive: it leaves the coefficients at all other <simple roots> unchanged, at least one of which is positive; a <root> has coordinates all of one sign. Repeating this height decrease reaches a <simple root>. Thus every <root> has squared length two in this connected single-bond system.
Write an arbitrary <positive root> as $a\alpha+b\beta+c\gamma$, with nonnegative integers $a,b,c$. Its squared length condition is
$$
a^2+b^2+c^2-ab-bc=1,\qquad
2=a^2+c^2+(a-b)^2+(b-c)^2.
$$
Thus $a,c\in\{0,1\}$. For $(a,c)=(0,0)$ the equation gives $b=1$; for $(1,0)$ or $(0,1)$ it gives $b=0$ or $1$; for $(1,1)$ it gives $b=1$. These are exactly the six coefficient triples in the displayed list. This proves completeness from the axioms, not merely from recognizing the <Dynkin diagram>.
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