This problem was incorrectly stated,it should be:
Suppose that $ a,b,c,d$ are natural numbers such that $ d|a^{2b} + c$ and $ d\geq a + c$.
Prove that
\[ d\geq a + \sqrt [2b]{a}
\]
Solution:
Suppose that $ d = a + k$,therefore:
\[ a + k|a^{2b} + c
\]
On the other hand $ a + k|a^{2b} - k^{2b}$ combining it with the given relation $ a + k|a^{2b} + c$,we conclude that
$ a + k|k^{2b} + c$.
Therefore,
$ \boxed{k^{2b} + c\geq a + k\geq a + c}$.
Finally
\[ k\geq\sqrt [2b]{a}
\]
But this means that $ d = a + k\geq a + \sqrt [2b]{a}$.