Finding the upper and lower bounds of a multivariable function

Question

I have the set $A = ${$r(a,b) = \frac{a+b}{a+b+3}, a \in [-1,1], b \in [-1,1]$}.

I started off by saying that $\frac{a+b}{a+b+3} < \frac{a+b}{a+b} = 1$. Thus, A is bounded from above by at least 1. Apparently that's wrong, but I don't understand why. What am I doing wrong, and how should I go about solving a problem like this? Thanks.

Answer 1

Because $a$ or $b$ could be negative. Rather, you can do it like

\begin{align*} \left|\dfrac{a+b}{a+b+3}\right|&=\left|1-\dfrac{3}{a+b+3}\right|\\ &\leq 1+\dfrac{3}{|a+b+3|}\\ &\leq 1+\dfrac{3}{3-|a+b|}\\ &\leq 1+\dfrac{3}{3-|a|-|b|}\\ &\leq 1+\dfrac{3}{3-1-1}\\ &=4, \end{align*} an upper bound for that is $4$, meanwhile a lower bound for that is $-4$, of course, this is a rough estimate.

Answer 2

First making

$$ a = \sin u, b = \sin v $$

we have

$$ r(a,b) \equiv r(u,v) = \frac{\sin u+\sin v}{\sin u+\sin v + 3} $$

but

$$ \sin u+\sin v = 2\sin(\frac{u+v}{2})\cos(\frac{u-v}{2}) \in [-2,2] $$

then finally

$$ \frac{-2}{-2+3} \le \frac{a+b}{a+b+3} \le \frac{2}{2+3} $$