Justifying $\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$ for large $x$ 
I need to find a way to justify the inequality $$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$$ that holds for large values of x.

As an instance, I will justify an example inequality just to show you the intended strategy. Consider the inequality
$$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5}{x}.$$ For example, $2x^2 - 7$ can be written as $x^2 + (x^2 - 7)$. Because $x^2 - 7$ is a positive value for all $x < -\sqrt{7}$, removing it from the denominator makes the absolute value of the fraction greater. Also note that when $x < -\frac{21}{10}$, the numerator $|5x + 21| < 5|x|$, and this happens for all $x < -\sqrt{7}$. So $$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5|x|}{x^2} = \frac{5}{x}$$ as long as $x < -\sqrt{7}$.

Here is a simple and direct way using just some rough estimates:
For $x>2$ you have surely
$$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} \stackrel{x>2}{<} \frac{5x^2 + 3x^2 + x^2}{x^3 - (x^2 +1)} = \frac{9x^2}{x^3 - (x^2 +1)}$$
Since you want to have a $10$ in the numerator, note that
$$\frac 1{10}x^3 > x^2+1 \Leftrightarrow x^2(\frac x{10}-1)>1$$
which is surely satisfied for $x>20$. Hence,
$$\frac{9x^2}{x^3 - (x^2 +1)} \stackrel{x>20}{<} \frac{9x^2}{x^3 - \frac 1{10}x^3} = \frac{10}x$$
Done.
