How does the Schwarz inequality prove the following? I don't quite see how the Schwarz inequality proves that, for 
$$f_n(x) = \frac{x}{1+nx^2}$$
we have 
$$|f_n(x)| \leq \frac{|x|}{2\sqrt{n}|x|} = \frac{1}{2\sqrt{n}}.$$
 A: The Cauchy-Schwarz (CS) inequality is used to get the inequality $1+nx^2 \geq 2\sqrt{n}x$. Namely, CS tells us $(1+nx^2)(1+1) \geq (1+\sqrt{n} x)^2$. Expand this, and you will get $1+nx^2 \geq 2\sqrt{n}x$.
Note. It is much easier to notice that $1+nx^2 - 2\sqrt{n} x = (1-\sqrt{n}x)^2\geq 0$. 
A: Consider the vectors $a = (1,x\sqrt{n})$ and $b = (x\sqrt{n},1)$. By the Cauchy-Schwarz Inequality: 
$$|a \cdot b| \le \|a\| \cdot \|b\|$$
$$|1 \cdot x\sqrt{n} + x\sqrt{n} \cdot 1| \le \sqrt{(1)^1+(x\sqrt{n})^2} \cdot \sqrt{(1)^1+(x\sqrt{n})^2}$$
$$|2x\sqrt{n}| \le 1+(x\sqrt{n})^2$$
$$2\sqrt{n}|x| \le 1+nx^2$$
Therefore, $|f_n(x)| = \dfrac{|x|}{1+nx^2} \le \dfrac{|x|}{2\sqrt{n}|x|} = \dfrac{1}{2\sqrt{n}}$, as desired.
It might be easier to use the AM-GM inequality, i.e. $\dfrac{1+nx^2}{2} \ge \sqrt{1 \cdot nx^2}$, i.e. $1+nx^2 \ge 2\sqrt{n}|x|$.
A: You can also show this using the fact that squares are nonnegative. $$(|x|\sqrt{n}-1)^2\geq 0$$
$$nx^2-2|x|\sqrt{n}+1\geq 0$$
$$nx^2+1\geq 2|x|\sqrt{n}$$
Therefore 
$$\frac{1}{nx^2+1}\leq \frac{1}{2|x|\sqrt{n}}$$
$$|f_n(x)|=\frac{|x|}{nx^2+1}\leq \frac{|x|}{2|x|\sqrt{n}}=\frac{1}{2\sqrt{n}}$$
