Let $m,q,v$ be integers with $m\geq 2$, and $v|q-1$. A certain result that I have which is not important for this question, holds when $$q^{\frac{m}{2}-2}(q-mv)\geq v^{m-1}. \quad (1)$$ I would like to have a clear idea about how large $q$ needs to be when $m$ grows large. Does it have to be roughly greater than $v^2$? Greater than $mv^2$ maybe? If this is a bit hand-wavy, please read on.
What I tried so far. First, we need to have $q>mv$, otherwise the left hand side of (1) would be negative and (1) would not hold. Hence, we can set $q-mv=cq$, for some $c$ with $0<c<1$. Then the condition can be written as $$cq^{\frac{m}{2}-1} \geq v^{m-1}$$ and solving for $q$ gives $$q \geq \left(\frac{1}{c}\right)^{\frac{2}{m-2}}v^{2+\frac{2}{m-2}}.$$ If $c$ was a constant, I could say that as $m$ approaches infinity, then the expression on the right hand side tends to $v^2$. So my conlcusion would be that when $m$ is large, then $q$ needs to be greater than $v^{2+\epsilon}$, for some $\epsilon=\epsilon(m)$. The problem is that $c$ is not really a constant, as it depends on $q,m,v$.
Am I right to be concerned about my observations above? If so, what can we say about how large $q$ should be with respect to $v$ and $m$, as $m$ grows large?