Let $(W, S)$ be a finite Coxeter system. Furthermore, let $V$ be a real vector space with a (finite) basis $\{ \alpha_s | s \in S \}$.
For every $s \in S$ one can define the reflection $\sigma_s : V \to V, v \mapsto v - 2 B(v, \alpha_s) \alpha_s $ where $B$ is the bilinear form defined by $B(\alpha_s, \alpha_t ) = - \cos(\frac{\pi}{m(s,t)})$.
There also exists a unique group homomorphism$\sigma : W \to \operatorname{GL}(V)$ such that $\sigma(s) = \sigma_s$, the so-called geometric representation of $(W,S)$. From now on we denote $\sigma(w) (v)$ for $v \in V, w \in W$ by just $w(v)$.
Now let $\Phi = \{w(\alpha_s) | w \in W, s \in S\} $ be the set of all roots and $\Pi = \{ v \in \Phi | v = \sum_{t \in S} c_t \alpha_t, c_t \geq 0 \forall t \in S\} $ the system of positive roots. Now consider $\alpha_s \in V$ for $s \in S$, a so-called simple root.
At this moment I am dealing with the proof of $s(\Pi \setminus \{ \alpha_s \}) = \Pi \setminus \{ \alpha_s \}$.
Let $\alpha \in \Pi \setminus \{ \alpha_s \}$. Then one can write $\alpha = \sum_{t \in S} c_t \alpha_t$ where the coefficients are non-negative and there exists a $t_0 \in S$ with $c_{t_0} > 0$.
Applying $s$ leads to $s(\alpha) = \sum_{t \neq s} c_t \alpha_t + \mu \alpha_s$ where I calculated $\mu = - c_s + 2 \sum_{t \neq s} c_t \cos(\frac{\pi}{m(s,t)}) $. Note that the cosines are non-negative since $m(s,t) \geq 2$ for $t \neq s$.
But I don't really know why $\mu \geq 0$ though since the $-c_s \leq 0$ could mess everything up.
It drives me insane that I could not find an answer yet. Could someone tell me why it still works? I really appreciate your help.
I'm not sure where this lemma lies in your study of root systems and whether I'm using facts that you're actually trying to prove, but here's a quick explanation. At least one of the $c_t$ is nonzero, otherwise $\alpha$ would be a multiple of $\alpha_s$ (other than $\pm 1$), which can't happen in a root system.
This means that $s(\alpha)$ has at least one simple root with positive coefficient, so $s(\alpha)$ is a positive root (all roots are either positive or negative). This forces $\mu \geq 0$.
Again, maybe you're trying to prove some of the facts that I've used.