Question: > 
> Show that the set $\{\frac{3x+2}{x^2+5}|x\in \mathbb{R}\}$ is bounded.
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A friend suggested that I should you Cauchy inequality to prove this. But I want to find a simpler way. Please help.

Answer: 
Let $$\displaystyle y = \frac{3x+2}{x^2+5}\Rightarrow yx^2+5y=3x+2$$

So we get $$yx^2-3x+(5y-2) = 0$$

Now for real values of $y,$ values of $x$ must be real.

So its $$\bf{Discriminant\geq 0}$$

So we get $$\displaystyle 9-4y\cdot (5y-2)\geq 0$$

So we get $$\displaystyle 9-20y^2+8y\geq0$$

So we get $$\displaystyle 20y^2-8y-9\leq 0\Rightarrow \left(y+\frac{1}{2}\right)\cdot \left(y-\frac{9}{10}\right)\leq 0$$

So we get $$\displaystyle -\frac{1}{2}\leq y\leq \frac{9}{10}$$
$$f(x)=\frac{3x+2}{x^2+5}$$
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:
$$\left|\frac{3x+2}{x^2+5}\right|\leq \frac{2}{5}+\frac{3|x|}{x^2+5} \leq \frac{2}{5}+\frac{3}{2\sqrt{5}}.$$