<s>[INST] Improper integral comparison test Having this integral
$$\int_1^{\infty}\frac{3x^2+2x +1}{x^3+6x^2+x+4}$$
In order to do the comparison test at some point it gets like
$$\frac{3x^2+2x +1}{x^3+6x^2+x+4}\geq \frac{1}{4x}$$
How is $\frac{1}{4x}$ found ? It doesnt seem obvious to me.
EDIT
Also how is this found
$$0\leq\frac{3x +1}{\sqrt{x^5+4x^3+5x}}\leq \frac{4}{x^{3/2}}$$
Where does 4 came from here ?Minimum value ? 3x1 +1  ? Why.
 [/INST] For positive $x$, the top is $\ge 3x^2$. For $x\ge 1$, the bottom is $\le x^3+6x^3+x^3+4x^3$.  
Edit: For your added question, the range of values of $x$ is not specified. However, if $x\ge 1$, then the top is $\le 4x$. The bottom is $\ge \sqrt{x^5}$. So for $x\ge 1$, the whole thing is $\le \frac{4x}{x^{5/2}}$, which simplifies to $\frac{4}{x^{3/2}}$. 
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