why $1+\sqrt{n}|x|^2 \ge 2\sqrt{n}|x|?$ why not $1+\sqrt{n}|x|^2 \ge 2\sqrt{n}|x|^2?$

Question

Is $f_n(x)=\frac{x}{1+nx^2}$ uniformly convergent on $\mathbb{R}?$

I found the answer here .

Here is an outline of the solution

$$\sup\limits_{x\in\mathbb R}|f_n(x)-0|=\sup\limits_{x\in\mathbb R}\dfrac{|x|}{|1+nx^2|}=$$ $$\sup\limits_{x\in\mathbb R}\dfrac{|x|}{1+\sqrt{n}|x|^2}\leq\sup\limits_{x\in\mathbb R}\dfrac{|x|}{2\sqrt{n}|x|}=\dfrac{1}{2\sqrt{n}}\overset{n\rightarrow\infty}{\longrightarrow }0$$ Therefore, $\sup\limits_{x\in\mathbb R}|f_n(x)-0|\rightarrow 0, $ i.e. $f_n$ converges uniformly to $0$.

My confusion : Im not getting why $1+\sqrt{n}|x|^2 \ge 2\sqrt{n}|x|?$ why not $1+\sqrt{n}|x|^2 \ge 2\sqrt{n}|x|^2?$

My thinking : If $1+\sqrt{n}|x|^2 \ge 2\sqrt{n}|x|^2$ then $f_n(x)$ will not uniformly convergent see Here

why it is a compulsory to take $1+\sqrt{n}|x|^2 \ge 2\sqrt{n}|x|?$

Answer 1

I think you have a typo.

You wrote $\sup\limits_{x\in\mathbb R}\dfrac{|x|}{|1+nx^2|}=\sup\limits_{x\in\mathbb R}\dfrac{|x|}{1+\sqrt{n}|x|^2}$

which isn't likely to be true as $|1+nx^2| = 1+nx^2 \ne 1 + \sqrt n |x|^2$.

If you replace this with $\sup\limits_{x\in\mathbb R}\dfrac{|x|}{|1+nx^2|}=\sup\limits_{x\in\mathbb R}\dfrac{|x|}{1+n|x|^2}$ your result will follow.

Bear in mind $1-2\sqrt n |x| +n|x|^2 = (1-\sqrt n|x|)^2 \ge 0$ so

$1+n|x|^2 \ge 2\sqrt n|x|$ (or use AM.GM) and you have your result.