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Let $0<b<a<1$. Does the following equality hold?

$b^{a-b} \geq \frac{b}{a}$

I feel that this can be proved by Bernoulli's inequality, but cannot figure out exactly how. Please help!

maomao
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1 Answers1

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Let $a - b = \epsilon \in (0,1)$. The inequality becomes $b^{\epsilon} \geq \frac{b}{b + \epsilon}$. By inverting both sides this becomes equivalent to $\left( \frac{1}{b} \right)^{\epsilon} \leq 1 + \frac{\epsilon}{b}.$

Now clearly $\frac{1}{b} > 1$ so Bernoulli's inequality with exponent in $(0,1)$ gives $$\left( 1 + \left(\frac{1}{b}-1\right) \right)^{\epsilon}\leq 1 + \epsilon \left(\frac{1}{b}-1\right) = 1 + \frac{\epsilon}{b} - \epsilon < 1 + \frac{\epsilon}{b}, $$

which is what we wanted.

Tanny Sieben
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