Proving $b^a(1-b)^{1-a} \leq a^a(1-a)^{1-a}$ where $0 \leq b \leq a \leq 1$

Question

I am trying to prove an inequality which boils down to the following: $$b^a(1-b)^{1-a} \leq a^a(1-a)^{1-a}$$ where $0 \leq b \leq a \leq 1$

Another equivalent formulation is: $$ a\log{b} + (1-a)\log{(1-b)} \leq a\log a + (1-a)\log (1-a)$$

How can I prove it?

Answer 1

For $x\in (0,1)$, define $$f_a(x)=a\log(x)+(1-a)\log(1-x)$$ Then $$f_a^\prime(x)=\frac a x -\frac {1-a}{1-x}=\frac {a-x} {x(1-x)}$$ So $f_a$ is maximum at $x=a$.