I am trying to find a nice way to verify that whenever $q \in (0,1)$ and $x \in (0,1)$, then $\frac{1-q}{q} - \frac{1-q^{x}}{xq^{x}} \geq 0 .$
Proving $\frac{1-q}{q} - \frac{1-q^{x}}{xq^{x}} \geq 0 $
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algebra-precalculus
inequality
2 Answers
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Take the derivative with respect to $q$ to get $\frac{q^{1-x} - 1}{q^2}$ which is negative for $q,x \in (0,1)$. This means that $\frac{1-q}{q} - \frac{1-q^x}{xq^x}$ is monotone decreasing in $q$ (for a chosen $x$), so it's enough to verify that the identity holds in $q=1$, which it does.
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Hint: Consider the function $f(x) = \frac{1-q^x}{x q^x},$ and express the inequality in terms of $f(x)$ and $f(1)$. Can you prove a property about $f$ that would imply the result?
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2It is a nice exercise to verify that (spoiler alert!) $f(x)$ is increasing on $(0,1]$, which implies that f(1) > f(x) for x < 1. (At some point you will actually need e^y > 1 + y to finish the proof.) – 2012-07-07