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# Help please!

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deleted.

Jun 10, 2019
edited by sinclairdragon428  Nov 20, 2019

### 2+0 Answers

#1
+6046
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$$\left(\dfrac 1 a + \dfrac 1 b\right)\left(\dfrac 1 b -\dfrac 1 a\right) = \dfrac{1}{b^2}-\dfrac{1}{a^2}$$

$$\text{to minimize this we want the largest possible magnitude of b\\and the smallest possible magnitude of a}\\ a = -1,~b=3\\ \dfrac{1}{3^2}-\dfrac{1}{(-1)^2} = \dfrac 1 9 - 1= -\dfrac 8 9$$

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Jun 10, 2019
#2
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If $$-5≤a≤-1$$ and $$1≤b≤3$$, what is the least possible value of $$\left(\dfrac{1}{a}+\dfrac{1}{b} \right) \left(\dfrac{1}{b}-\dfrac{1}{a} \right)$$?

1.)

$$\left(\dfrac{1}{a}+\dfrac{1}{b} \right) \left(\dfrac{1}{b}-\dfrac{1}{a} \right) = \dfrac{1}{b^2}-\dfrac{1}{a^2}$$

2.)

$$\begin{array}{lrcll} &\mathbf{ -5}\le &\mathbf{a}& \le \mathbf{-1} \quad | \quad square \\ & 25\le &a^2& \le 1 \\ (1) & \mathbf{\dfrac{1}{25}} \ge & \mathbf{\dfrac{1}{a^2}} & \ge \mathbf{1} \\ \hline & \mathbf{1} \le &\mathbf{b}& \le \mathbf{3} \quad | \quad square \\ & 1 \le &b^2& \le 9 \\ (2)& \mathbf{1} \ge & \mathbf{\dfrac{1}{b^2}} & \ge \mathbf{\dfrac{1}{9}} \\ \hline (2)-(1): & \mathbf{1} -\mathbf{\dfrac{1}{25}} \ge & \mathbf{\dfrac{1}{b^2}} -\mathbf{\dfrac{1}{a^2}} & \ge \mathbf{\dfrac{1}{9}} -\mathbf{1} \\ & \dfrac{24}{25} \ge & \dfrac{1}{b^2} - \dfrac{1}{a^2} & \ge -\dfrac{8}{9} \\ \end{array}$$

The least possible value is $$\mathbf{-\dfrac{8}{9}}$$

Jun 10, 2019