I run into this inequality

$$ (a + b)^{1 - \epsilon} \;a < b $$

where $a \in \mathbb{Z}^+$ and $\epsilon \in (0, 1)$. What value (w.r.t $a$ and $\epsilon$) should I set $b$ equal to such that this inequality holds for all $a \in \mathbb{Z}^+$? (if possible).

What I have gotten so far:

$$ \frac{a + b}{(a + b)^{\epsilon}} \; a < b $$ $$ \Downarrow $$ $$ a^2 + ab < (a + b)^{\epsilon} \; b \leq (a^{\epsilon} + b^{\epsilon}) b $$ $$ \Downarrow $$ $$ a^2 + ab < a^{\epsilon}b + b^{1 + \epsilon} $$

I am struggling to proceed from here.