Improper integral comparison test

Question

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.

Answer 1

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 $\frac{4}{x^{3/2}}$.

Answer 2

For example, and taking into account the fact that for large values $\;x\;$ only the summands with the largest exponent of $\;x\;$ are important (in this sense), we get

$$\frac{3x^2+2x +1}{x^3+6x^2+x+4}\ge\frac{3x^2}{x^3+6x^3+x^3+4x^3}=\frac1{4x}$$