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

Question

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

Answer 1

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.

Answer 2

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?