How to prove the equivalence between the two statements of ABC conjecture? The ABC conjecture stated by wikipedia says the following statements are equivalent:
I.
For $\epsilon>0$, there are finite coprime triple $(a,b,c)$ satisfying $a+b=c$ such that $\mathrm{rad}(abc)^{1+\epsilon}<c$
II.
For  $\epsilon>0$, there exists $C_\epsilon>0$,such that for all coprime triple $(a,b,c)$ satisfying $a+b=c$, $C_{\epsilon}\mathrm{rad}(abc)^{1+\epsilon}>c$ holds.
How to prove $II \implies I$?
thanks.
 A: *

*If $C_\epsilon < 1$ then the implication is obvious. 

*Consider $\epsilon' = \epsilon / 2$. There exists $C_{\epsilon'}$ such that for all coprime triples with $a+b = c$, the statement 
$$ \mathrm{rad}(abc) > \frac{c^{1/(1+\epsilon')}}{C_{\epsilon'}^{1/(1+\epsilon')}} \tag{*}$$ is true by the ABC conjecture version (II). Now, 
$$ \frac{1}{1+\epsilon'} = \frac{1}{1+\epsilon} + \frac{\epsilon}{(1+\epsilon)(2+\epsilon)} $$
So if for $c$ large enough (such that $\exp\left( \frac{\epsilon}{2+2\epsilon}\log c \right) > C_{\epsilon'}$ is true), (*) implies that 
$$ \mathrm{rad}(abc) > \frac{c^{1/(1+\epsilon')}}{C_{\epsilon'}} = c^{1/(1+\epsilon)} \cdot \left(\frac{c^{\epsilon/(2+2\epsilon)}}{C_{\epsilon'}} \right)^{1/(1+\epsilon')} \geq c^{1/(1+\epsilon)}$$
which is precisely the statement of ABC conjecture version (I). Now using that $C_{\epsilon'}$ is a finite number, there are only finitely many $c$ for which 
$$ \exp\left( \frac{\epsilon}{2+2\epsilon}\log c \right) \leq C_{\epsilon'} $$
is true, and these finitely many $c$s account for the finitely many possible exceptions of ABC conjecture version (I). 

The reverse implication is also simple. Let $C_\epsilon$ be defined as 
$$ C_\epsilon = 1 + \sup \frac{c}{\mathrm{rad}(abc)^{1+\epsilon}} $$
If (I) holds, then this sup is taken over a finite number of cases where the ABC inequality is violated, and so is finite. 
