Least Positive integer satisfying the condition $x^{2} -4x > \cot^{-1}x$

Question

What is the least positive integer satisfying the condition: $$x^{2} -4x > \cot^{-1} (x)$$

Graphically entering the value $y=x^{2} -4x$ and $y= \cot^{-1}(x)$, I concluded that the answer is $x=5$ as it satisfies the condition of a positive integer, but I would like to solve this question using a mathematical equation if possible.

Answer 1

Let $$ f(x)=x^2-4x-\cot^{-1}x. $$ Note that $$ f'(x)=2x-4+\frac{1}{x^2+1}>0 $$ implies $f(x)$ is increasing for $x\ge 2$. Clearly when $x=1,2,3,4$, the inequality does not hold since the LHS is non-positive. When $x=5$, the inequality holds. Thus $x=5$ is the smallest integer satisfying the inequality.

Answer 2

$x \in \mathbb{Z^+} \Rightarrow cot^{-1} x > 0.\;\;$ 5 is the smallest positive integer $x$ such that $x^2 - 4x > 0.$