<s>[INST] find an $A$ so that for all $n \ge A (2^n+3^n)/(4^n+5^n)/(1/1000)$ 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 
 [/INST] For the similar problem, the inequality is false.  If we take the base $10$ log of each side, we get $n \log 1.01 - 147 \log n \lt 2$ or $ .00432n - 147 \log n \lt 2$  Now we just need to take $n$ large enough.  $n=10^6$ is easily enough.
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