Help with inequality: arithmetic vs weighted geometric

Question

Let $p$, $q$ be positive real numbers such that $p+q < 1$. Prove that

$$ \frac{p+q}{2} \leq \left( p^p q^q \right)^{1/(p+q)} $$

I'm not sure the assumption $p+q < 1$ is really necessary. Anyhow, I can't think of a nice way to prove it.

Answer 1

Equivalently, we want to prove $$\dfrac{1}{2} \leq (\dfrac{p}{p+q})^{\dfrac{p}{p+q}}(\dfrac{q}{p+q})^{\dfrac{q}{p+q}}$$ i.e. $$\log\dfrac{1}{2} \leq \dfrac{p}{p+q}\log(\dfrac{p}{p+q}) + {\dfrac{q}{p+q}}\log(\dfrac{q}{p+q})$$

Consider the function $$f(x) = x\log x + (1-x)\log(1-x), x\in (0,1)$$ we have $$f'(x) = 1 + \log x- 1 -\log(1-x) = \log\dfrac{x}{1-x}$$

thus $f'(x) \leq 0$ when $x \in (0,\frac{1}{2}]$ and $f'(x) \geq 0$ when $x \in [\frac{1}{2},1)$

So $f$ gets its minimum when $x = \dfrac{1}{2}$ and the minimum is equal to $\log\frac{1}{2}$, then $f(x) \geq \log\frac{1}{2}$. In particular $f(\dfrac{p}{p+q}) \geq \log\frac{1}{2}$.