How fast would a nondeterministic algorithm for an EXPTIME-complete problem have to be to imply $P \neq NP$?
I think its easier to turn it around.

If $\mathsf{P}=\mathsf{NP}$, then $\mathsf{NTIME}(T(n)) \subset \mathsf{DTIME} ((T(n))^c)$ 
for some constant $c$, and any $T(n) &gt; n$.
Since $\mathsf{DTIME} ((T(n)^c)$ does not contain 
$\mathsf{DTIME}(T(n)^{c}\log T(n)) \subset \mathsf{DTIME}(T(n)^{c+1})$, 
this means we cannot solve, say all problems in $\mathsf{DTIME}(2^n)$ in 
$\mathsf{NTIME} (2^{\epsilon n})$ for some $\epsilon$.
So a non-deterministic time $2^{o(n)}$ algorithm for a problem complete for
$\mathsf{DTIME} (2^n)$ under quasi-linear reductions would be enough to prove $\mathsf{P} \neq \mathsf{NP}$.