<s>[INST] bound on $\biggr\rvert \frac{x+2y-2}{x^2+y^2+1} \biggr\lvert $? Question: How to get a good bound on $\biggr\rvert \frac{x+2y-2}{x^2+y^2+1} \biggr\lvert $?
Context: I want to show that $\frac{x+2y-2}{x^2+y^2+1}$ attains maximum and minimum value on $\Bbb R^2$. So I need to find some compact set $K$ which contains the minimum and maximum, and show that on $\Bbb R^2 - K$ , the absolute value of the function is small.
Attempt:
$\biggr\rvert \frac{x+2y-2}{x^2+y^2+1} \biggr\lvert 
\leq \frac{|x+2y-2|}{x^2+y^2+1} \leq \frac{\sqrt{x^2+y^2}+2(\sqrt{x^2+y^2})-2}{x^2+y^2+1} \leq \frac{3\sqrt{x^2+y^2}-2}{x^2+y^2}
$
This is good enough to bound the function when $x^2+y^2$ is large, but is there a better way?
But I am not sure how to get rid of the $-2$ in the numerator.
 [/INST] How about 
$$\left| \dfrac {x+2y-2}{x^2+y^2+1} \right| \le \dfrac {|x|+|2y|+|2|}{|x|+|y|+1} \le \dfrac {2|x|+2|y|+2}{|x|+|y|+1} = 2$$
This should hold when $|x|, |y| > 1$.
You can take $K$ to be the unit square. You can check that $f\left(- \dfrac 14, - \dfrac 12 \right) < -2$, so indeed the function has an absolute minimum.
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