If -5 <= a <= -2 and 2 <= b <= 5, what is the least possible value of $\displaystyle\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}-\frac{1}{a}\right)$? Express your answer as a common fraction.
+876 Well, let's see.
We can expand and simplify the product: $\left(\frac{a + b}{ab}\right)\left(\frac{a - b}{ab}\right) = \frac{a^2 - b^2}{a^2 b^2}$
We could try and make a negative with them, but $b$ is always positive, so that doesn't work.
Because of the squares, the numerator will always be nonnegative(the least value is $(-2)^2 - 2^2 = 0$) and the denominator always positive. So the least positive value is when the numerator is 0, which gives 0 as the solution.
