Proof that $ \frac{ln{x}}{(x^3-1)} <\frac{x}{x^3} , \forall x \in[2,\infty) $

Question

$$ \frac{ln{x}}{(x^3-1)} <\frac{x}{x^3} , \forall x \in[2,\infty) $$

This is specifically for an improper integral question, where the left term needs to be proven convergent or divergent for the interval $$ [2,\infty) $$

Answer 1

Hints:

$$x^3-1=(x-1)(x^2+x+1)$$

$$\ln(x)<x-1$$

$$x^2+x+1>x^2$$

(the last one for $x\ge2$)

Answer 2

Let $$f(x)=\ln x-x+\frac{1}{x^2} \Rightarrow f'(x)=\frac{1}{x}-1-\frac{2}{x^3}<0, ~\mbox{for}~x\ge 1. $$ So $f(x)$ is decreasing function on $[1,\infty].$ This means that $f(x) \le f(1) \Rightarrow f(x) \le 0$. Rearrangin this we get $$\frac{\ln x}{x^3-1} \le \frac{x}{x^3},~~ x > 1.$$