If $\frac{1 - \phi_1}{\phi_2} > 1$, $\frac{1 + \phi_1}{\phi_2} > 1$ and $|\phi_2| < 1$, how can I proceed from here to show that $\phi_1 + \phi_2 < 1$, $\phi_2 - \phi_1 < 1$ and $|\phi_2| < 1$?
I can show this by considering two cases such as $-1 <\phi_2 < 0$ and $0 <\phi_2 < 1$. But is there an elegant way to solve this.
Hint: adding the two inequalities gives $\require{cancel}\,\frac{1 - \cancel{\phi_1}}{\phi_2}+\frac{1 + \cancel{\phi_1}}{\phi_2} \gt 2 \iff \frac{2}{\phi_2} \gt 2\,$, so $\,\phi_2 \gt 0\,$. In fact, it follows that $\,\phi_2 \in (0,1)\,$, so the condition $\,|\phi_2| \lt 1\,$ is redundant altogether.
Then the given inequalities imply that $\,\phi_2 -1 \lt \phi_1 \lt 1 - \phi_2\,$.