<s>[INST] 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 $x < -\sqrt{7}$.
 [/INST] mrsamy's method is quite elegant, though it says nothing about how large $x\gg 1$ should be.
If you need a thighter inequality, the best way is to develop
$$f(x)=(5x^2+3x+1)x-10(x^3-x^2-1)$$
Graph it and notice it is negative for some $x$ close to (and below) $3$
Then just calculate $f(u+3)=-5u^3-32u^2-56u-5<0\quad$ for $\ u\ge 0$
Therefore the inequality is valid as soon as $x\ge 3$ (the actual value is $\approx 2.9057$).
</s>