Suppose that I have these two inequalities over the real domain $x>0$ and the parameter $a>0$; here, $f(x)$ and $g(x)$ are real functions
$$ -|g(x)|+1+a \; \leq f(x) \leq \; |g(x)|+1+a \qquad (1) $$ $$ f(x) > \; |g(x)|+1 \qquad\qquad\qquad\qquad\qquad\qquad (2) $$
My question Under what constraint over the functions $f(x)$ and $g(x)$, I can claim that there always exist $x$'s which satisfy both $(1)$ and $(2)$? In other words, under what conditions, I can claim the inequalities $(1)$ and $(2)$, are always, have an intersection part?
Note that the second inequality in $(1)$ together with $(2)$ imply that $f(x)-|g(x)|\in(1,1+a]$, i.e. $f(x)=|g(x)|+1+\delta$ for some $0<\delta\leq a$.
Putting this into the first inequality in $(1)$ implies $-|g(x)|+1+a\leq |g(x)|+1+\delta$, i.e. $|g(x)|\geq \frac{a-\delta}{2}$. Note that the smaller $|g(x)|$ is, the larger $\delta$ has to be to allow for this to happen. Specifically $\delta$ has to satisfy $\delta> \max{(a-2|g(x)|,0)}$.
Therefore, given arbitrary $g$, you can define $f(x):=1+|g(x)|+h(x)$ for an arbitrary strictly positive function $h$ such that $a-2|g(x)|\leq h(x)\leq a$ and you will always satisfy $(1)$ and $(2)$.