Proving $x^x$ is strictly less than $(1-x)^{1-x}$ on $(0,\frac{1}{2})$

Question

I am trying to prove the following:

$x^x < (1-x)^{1-x}$ for $x \in (0,\frac{1}{2})$

(or take logs of both sides to make it nicer if you want)

I can draw pictures to convince myself this is true, but I'm wondering how to structure a formal argument. It feels that the second derivative being strictly decreasing in $x$ should help, but I can't quite piece together an argument.

Any ideas would be gratefully received!!

Answer 1

Let $f(x)=(1-x)\ln(1-x)-x\ln{x}.$

Thus, since $$f''(x)=\frac{1}{1-x}-\frac{1}{x}=\frac{2x-1}{x(1-x)}<0,$$ $$f\left(\frac{1}{2}\right)=0$$ and $$\lim_{x\rightarrow0^+}f(x)=0,$$ we obtain $$f(x)>0$$ or $$x^x<(1-x)^{1-x}.$$

Answer 2

Well we have $x < 1-x$ on that range so both the base and the exponent of $(1-x)^{1-x}$ are larger than the base and exponent of $x^x$ so we can do this in a few steps by noting $$x^x < (1-x)^x < (1-x)^{1-x}$$ because for positive $x$ and some $c > 0$ we know both c^x and $x^c$ are strictly monotonically increasing.

Edit: I overlooked that $c^x$ is decreasing when $0<c<1$ so the argument fails as pointed out in the comments.