Show that $f_{n}(x)=\frac{x}{1+nx^{2}}$ converges uniformly. 
Show that $f_{n}(x)=\frac{x}{1+nx^{2}}$ converges uniformly.

I was looking at Rudin's answer to this proof:

I don't understand the part I squared in red. How do I use the Cauchy-Schwarz inequality to get this result? I am only used to using the inequality when dealing with converging series.
 A: Refer to Cauchy-Schwarz inequality:
$$(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2.$$
In your case:
$$(1+nx^2)^2=(1+n|x|^2)(n|x|^2+1)\ge (\sqrt{n}|x|+\sqrt{n}|x|)^2=(2\sqrt{n}|x|)^2.$$
You can also use AM-GM:
$$a^2+b^2\ge 2ab \ \text{or} \ \ |a|+|b|\ge 2\sqrt{|ab|}.$$
In your case:
$$1+nx^2\ge 2\sqrt{1\cdot nx^2}=2\sqrt{n}|x|.$$
And finally note that when denominator $1+nx^2$ decreases to $2\sqrt{n}|x|$, the value of fraction increases:
$$|f_n(x)|=\left|\frac{x}{1+nx^2}\right|=\frac{|x|}{|1+nx^2|}\le \frac{|x|}{|2\sqrt{n}|x||}=\frac{|x|}{2\sqrt{n}|x|}=\frac{1}{2\sqrt{n}}.$$
A: You can think of it as such:
$$1 - 2|x|\sqrt{n} + n|x|^2 =(1-|x|\sqrt{n})^2 \ge 0$$
No Cauchy -Schwarz is needed.
A: Trying to reconstruct it - we could apply Cauchy-Schwarz to say that
$$2\sqrt{n}|x| =\langle (1,\sqrt{n}|x|),(\sqrt{n}|x|,1)\rangle \le 1+nx^2$$
and then solve to $\frac{|x|}{1+nx^2} \le \frac1{2\sqrt{n}}$.
Of course, the equally classical AM-GM inequality is also easy to apply here; $1+nx^2 \le 2\sqrt{1\cdot nx^2}=2\sqrt{n}|x|$. Or we could do that by writing it as the square of a difference. When you get down to two variables, a lot of those inequalities converge to the same thing.
