I'm studying root systems and coming up with the following observation.

Let $\Phi$ be an irreducible root system and $\Phi^+$ be a positive root system. Denote by $\Delta=\{\alpha_1, \alpha_2,\ldots,\alpha_n\}$ be the corresponding base. For $\beta, \gamma \in \Phi^+$, $\alpha_1, \alpha_2 \in \Delta$ with the properties that $\beta=\gamma +\alpha_1+\alpha_2$, $\gamma +\alpha_1 \in \Phi^+, \gamma +\alpha_2\in \Phi^+, \alpha_1+ \alpha_2 \notin \Phi^+$ then

Statement: There exists $\alpha_3 \in \Delta\setminus\{\alpha_1,\alpha_2\}$ such that either $\gamma +\alpha_1-\alpha_3 \in \Phi^+$ or $\gamma +\alpha_2-\alpha_3 \in \Phi^+$.

By a quick looking at lists of all irreducible root systems, I personally think that the above statement can be correct. I tried to make some effort in order to find a uniform proof for that. Suppose to the contrary that, $\gamma +\alpha_1-\alpha_i \notin \Phi^+$ for all $\alpha_i \in\Delta\setminus\{\alpha_1\}$ and $\gamma +\alpha_2-\alpha_j \notin \Phi^+$ for all $\alpha_j \in\Delta\setminus\{\alpha_2\}$. It follows that, the standard inner product $(\cdot,\cdot)$ draws some relations between roots that $(\gamma +\alpha_1,\alpha_i) \le 0$ for all $\alpha_i \in\Delta\setminus\{\alpha_1\}$ and $(\gamma +\alpha_2,\alpha_j) \le 0$ for all $\alpha_j \in\Delta\setminus\{\alpha_2\}$. I'm aware of the event that positive roots forming pairwise obtuse angles must be linearly independent. So we have 2 independent sets having the same cardinality with $\Delta$ are $\{\gamma +\alpha_1,\alpha_2, \ldots,\alpha_n\}$ and $\{\alpha_1,\gamma +\alpha_2, \ldots,\alpha_n\}$. Can this lead to any contradiction? For if not, could you disprove the statement by some counter-example?

Any help would be much appreciated.