Let $\Delta$ be an indecomposable root system in a real inner product space $E$, and suppose that $\Phi$ is a simple system of roots in $\Delta$, with respect to an ordering of $E$. If $\Phi = \{\alpha_1, \ldots, \alpha_l \}$, prove that $$ \alpha_1 + \cdots + \alpha_l \in \Delta $$
I know that any positive root $\gamma$ may be written as a sum of simple roots, and furthermore that every partial sum is itself a root, but I am unsure if that will help me or not. Any hints to get me started?
Edit: I had an idea since I posted but am now stuck on the proof. We know that $(\alpha_i, \alpha_j) \leq 0$ for $i \neq j$. We also know that if $(\alpha_i,\alpha_j) < 0$, then $\alpha_i + \alpha_j$ is a root. My thoughts are an induction-style proof: If $(\alpha_1,\alpha_2) < 0$, then $\alpha_1+\alpha_2 \in \Delta$. Assume that $\alpha_1 + \cdots + \alpha_{l-1}$ is a root then $$ (\alpha_1 + \cdots + \alpha_{l-1}, \alpha_l) = (\alpha_1, \alpha_l) + \cdots + (\alpha_{l-1},\alpha_l) < 0 $$ Hence, $$ \alpha_1 + \cdots + \alpha_l \in \Delta $$ However, I am not sure how to prove these are strict inequalities. I assume it has to do with the fact that $\Phi$ is indecomposable, but I am unsure. Any ideas?
First, I would let $\Phi$ be a base for an indecomposable root system, and choose labels for the elements of $\Phi$ according to the following algorithm:
Suppose you have chosen $\Phi_k:=\{\alpha_1,\ldots,\alpha_k\}$ such that the corresponding root system $\Delta_k$ is indecomposable (when $k=1$ this is automatic). Then, there exists $\alpha_{k+1}\in\Phi$ such that $\Phi_{k+1}$ is a base for an indecomposable root system $\Delta_{k+1}$. Indeed, otherwise $$\mathrm{span}(\Phi_k)\perp \mathrm{span}(\Phi\backslash\Phi_k),$$ so $\Delta$ is not indecomposable, a contradiction.
This is enough to obtain strict inequality since, by construction, we must have $(\alpha_j,\alpha_l)<0$ for some $j<l$.