<s>[INST] Evaluating product of exponent and polynomial In a probability theory problem, I need to solve an inequality over $n\in\mathbb{N}$ which can be expressed in a general form like this: 
$a^n ( b_1 n^2 + b_2 n + b_3) \leq c$ where $a, b_1, b_2, b_3, c$ are real numbers.
What is the standart way (or a good way) to tackle this? Is there an analytic solution?
For reference, here is the exact inequality:
$0.8^n ( \frac{1}{32} n^2 - \frac{7}{32} n + 1) \leq 0.1$
 [/INST] Let's clean up a bit
$$ 0.8^n\left(\frac{1}{32}n^2-\frac{7}{32}n+1\right)\leq 0.1 $$
$$ \left(\frac{8}{10}\right)^n\left(\frac{1}{32}n^2-\frac{7}{32}n+1\right)\leq \frac{1}{10} $$
$$ \frac{8^n}{10^n}\left(\frac{1}{32}n^2-\frac{7}{32}n+1\right)\leq \frac{1}{10} $$
$$ \frac{8^n}{8}\left(\frac{1}{4}n^2-\frac{7}{4}n+8\right)\leq \frac{10^n}{10} $$
$$ 8^{n-1}\left(\frac{1}{4}n^2-\frac{7}{4}n+8\right)\leq 10^{n-1} $$
$$ \frac{1}{4}n^2-\frac{7}{4}n+8\leq \frac{10^{n-1}}{8^{n-1}} $$
$$ \frac{1}{4}\left(n^2-7n+32\right)\leq \left(\frac{5}{4}\right)^{n-1} $$
$$ n^2-7n+32\leq 4\left(\frac{5}{4}\right)^{n-1} $$
Now let's find the minimum of the quadratic function 
$$ \frac{d}{dn}\left[ n^2-7n+32\right]=2n-7=0 $$
$$ n=\frac{7}{2}=3.5 $$
However since $n\in\mathbb{N}$ then you can start with $n=4$ and work your way up from there. I would choose $n$ in increments of $10$ as it usually doesn't take long before an exponentially increasing function becomes greater than a quadratic function. After a few more than $3$ computations, you'll realize that $n\geq 21$ renders the inequality true.
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