# Proving $\frac{1-q}{q} - \frac{1-q^{x}}{xq^{x}} \geq 0$

I am trying to find a nice way to verify that whenever $q \in (0,1)$ and $x \in (0,1)$, then $$\frac{1-q}{q} - \frac{1-q^{x}}{xq^{x}} \geq 0 .$$

-
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.
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?
It 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.) –  TMM Jul 7 '12 at 23:26