= Solution
There is an exception to the printed assertion: for $\beta=-\alpha$, the <inner product> is negative but $\alpha+\beta=0$ is not a <root>. This already occurs in the rank-one <root system> $R=\{\alpha,-\alpha\}$.
For the intended assertion, assume $\beta\ne-\alpha$. By the preceding list, at least one of $n_{\alpha,\beta},n_{\beta,\alpha}$ equals $-1$. If the first does, the reflection axiom gives
$$
s_\alpha(\beta)=\beta-n_{\alpha,\beta}\alpha=\beta+\alpha\in R.
$$
If the second does, use $s_\beta(\alpha)=\alpha+\beta$ instead. Thus the precise reusable conclusion is
$$
\boxed{(\alpha,\beta)<0,\ \beta\ne-\alpha\quad\Longrightarrow\quad\alpha+\beta\in R.}
$$
This proves that <obtuse nonopposite roots have a root sum> directly from the reflection and integrality axioms, without appealing to the classification of <root systems>.
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