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What are the (possible) minimum and maximum values of the rational algebraic expression

$$\frac{AB}{(A + B)(A - B)},$$

if $A, B \in \mathbb{R}$ with $B < A$?

Thank you!

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    $\begingroup$ It might help to note that $\displaystyle \frac{AB}{(A+B)(A-B)} = \frac{AB}{A^2 - B^2}$ $\endgroup$
    – amWhy
    Mar 4, 2013 at 1:55
  • $\begingroup$ Thank you @André Nicolas and Avi Steiner. If only I could accept two answers, I would. I am accepting the more comprehensive answer by Avi for now. $\endgroup$ Mar 5, 2013 at 13:31

2 Answers 2

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Either $B$ or $A$ must be non-zero (otherwise the given expression is undefined). If one of them is equal to zero, the whole expression is zero, which is clearly not a max or a min. Thus, we may assume that both $B$ and $A$ are non-zero.

Set $r=B/A$. Then $$\frac{A^2-B^2}{AB} = \frac{A}{B} - \frac{B}{A} = r - \frac{1}{r}.$$ Thus, since $r$ can be any real number (except $0$), our problem reduces to finding the max and min of this expression (or rather, of the reciprocal of this expression). If $r\leq1$, then $$r-\frac{1}{r} \leq r-1<r,$$ so since $r$ can be as small as we want, the given expression has no minimum. Similarly, if $r\geq1$, then $$r-\frac{1}{r} \geq r-1,$$ so since $r$ can be as large as we want, the given expression has no maximum.

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There is no maximum. By choosing, for example, $B=1$, and $A$ a tiny bit bigger than $1$, we can make our expression larger than any prespecified real number.

There is also no minimum.By choosing $A=-1$, and $B$ a tiny bit less than $-1$, we can make our expression less than any prespecified real number.

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