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 $$\displaystyle -\frac{1}{2}\leq y\leq \frac{9}{10}$$
When $|x| \ge 3$ we have $|3x+2|\le 3|x|+2 \le (x.x+2)<(x.x+5)=|x^2+5|\ne 0 $ so the absolute value of the ratio is less than $1$. When $|x|<3$ we have $|3x+2|\le 3|x|+2<3.3+2=11$ , and $x^2 + 5 \ge 5$ so the absolute value of the ratio is less than $11/5.$ For this Q any bound less than infinity will do.