So $1000(2^n+3^n)\le 4^n+5^n$. I take $1000\cdot2^n\le 4^n$ and $1000\cdot3^n\le5^n$ so that adding both gives the inequality, theorem in the ordered fields. So $1000\cdot2^n\le2^2n$. This leads me to an $A$ for $n = 10$ as $1000\cdot2^n < 2^n\cdot2^n$ so $1000\leq2^n$ for $n=10$. But this does not count for $1000\cdot3^n<5^n$. I think to see that $1000\cdot3^n<3^n\left(\frac{5}{3}\right)^n$ so $1000<\left(\frac{5}{3}\right)^n$. so $n = \log_{\frac{5}{3}} (1000)$. This leads me to an $A = 14$ which is my final answer.
Any other suggestions.
A similar problem is $\displaystyle\frac{(1.01)^n}{n^{147}} < 100$. This is tougher. Any hints