Prove that $x ^ {1 - x} (1 - x) ^ {x} \le \frac {1}{2}$

Question

How do I prove $x ^ {1 - x}(1 - x) ^ {x} \le \frac{1}{2}$, for every $x \in (0, 1)$.

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For $x = \frac {1}{2}$ the LHS is equal to one half. I tried studying what happens when $x \lt \frac {1}{2}$ and its correspondent, but to no result.

Answer 1

Because $1-x+x=1$, and $x \in (0,1)$, by concavity of the $\ln$, one has $$(1-x)\ln(x)+x \ln(1-x) \leq \ln((1-x)x+x(1-x))=\ln(2x(1-x))$$

But it is easy to see that for every $x$, $$x(1-x) \leq \frac{1}{4}$$ so $$(1-x)\ln(x)+x \ln(1-x) \leq \ln \left( \frac{1}{2}\right)$$

Now take the $\exp$ : you get directly $$x^{1-x}(1-x)^x \leq \frac{1}{2}$$

Answer 2

let $a=x,b=1-x$,

$a+b=1$,

By AM-GM$$\frac{1}{2}=\frac{{(a+b)}^2}{2}\ge 2ab=\frac{ab+ba}{a+b}\ge \sqrt[a+b]{a^b b^a}=a^bb^a=x^{1-x}{(1-x)}^{x}$$

Answer 3

Let $$ f(x)=\ln[x ^ {1 - x}(1 - x) ^ {x}]=(1-x)\ln x+x\ln(1-x) $$ and then $$ f'(x)=-\ln x+\frac{1-x}{x}+\ln(1-x)-\frac{x}{1-x}, f''(x)=-\frac{1-x+x^2}{x^2(1-x)^2} .$$ Clearly $x=\frac12$ is the only point in $(0,1)$ such that $f'(x)=0$ since $f''(x)<0$ in $(0,1)$. Thus $f(x)$ obtains the max at $x=\frac12$, namely $$ f(x)\le \ln(\frac12). $$ So $$ x ^ {1 - x}(1 - x) ^ {x}\le \frac12. $$