<s>[INST] Show that the range of the function $f(x)=(3x+2)/(x^2+5)$ is bounded 
Show that the set $\{\frac{3x+2}{x^2+5}|x\in \mathbb{R}\}$ is bounded.

A friend suggested that I should you Cauchy inequality to prove this. But I want to find a simpler way. Please help.
 [/INST] $$f(x)=\frac{3x+2}{x^2+5}$$
is a continuous function on $\mathbb{R}$ for which
$$ \lim_{x\to \pm\infty}f(x) = 0,$$
hence $f$ is bounded. By computing $f'(x)$, we may check that the stationary points of $f(x)$ occur at $x=-3$ and $x=\frac{5}{3}$. If we compute the values of $f(x)$ at such points, we get:
$$-\frac{1}{2}\leq f(x) \leq \frac{9}{10}.$$
Anyway, the Cauchy-Schwarz inequality is a pretty fast way to go:
$$ \left|3x+2\right| = \left|3x+\frac{2}{\sqrt{5}}\cdot\sqrt{5}\right|\leq \sqrt{9+\frac{4}{5}}\sqrt{x^2+5} $$
hence:
$$ \left| f(x)\right |\leq \frac{7}{\sqrt{5}\sqrt{x^2+5}}\leq \frac{7}{5}.$$
The AM-GM plus the triangular inequality work pretty well, too:
$$\left|\frac{3x+2}{x^2+5}\right|\leq \frac{2}{5}+\frac{3|x|}{x^2+5} \leq \frac{2}{5}+\frac{3}{2\sqrt{5}}.$$
</s>