Show that $x^x(1-x)^{(1-x)} \geq \frac{1}{2}$ for $x \in (0,1)$.

Question

My approach:

First assume $x$ is a rational. Let $x=\frac{a_1}{b_1}$ and $1-x = \frac{a_2}{b_2}$, where $a_1,a_2,b_1,b_2 \in \mathbb{Z}^{+}, a_1 < b_1, a_2<b_2.$

Somehow show that $$\left(\frac{a_1}{b_1}\right)^{\left(\frac{a_1}{b_1}\right)} \left(\frac{a_2}{b_2}\right)^{\left(\frac{a_2}{b_2}\right)} \geq \frac{1}{2}$$ using the AM-GM inequality. Then by the continuity of $f(x) = x^x(1-x)^{1-x}$ on $(0,1)$, we can show that $\forall x \in (0,1), f(x) \geq \frac{1}{2}.$

I know that cleverly applying AM-GM is the non-trivial part :)

Answer 1

There's a one-line weighted AM-GM solution.

Hint: It seems like the LHS is "product of several terms", which is usually the GM side. So we can't quite work with this directly. How do we make the RHS the "product of several terms (maybe raised to some power)", while turning the LHS into "sum of several terms (maybe raised to some power)"?
Well, we ...

Multiply throughout by $ 2x^{1-x} ( 1-x)^ x$.
This gives us:

$$ 2x(1-x) \geq x^{1-x} (1-x)^x $$

which is clearly true by weighted AM-GM.
The terms should be obvious, especially on the GM side, so just check that they have the appropriate sum. Note that the sum of the weights is 1.