If $a,b\in [0,1], a(1-a) \leq b(1-b)$, is it necessary that $-a\log_2 a - (1-a)\log_2 (1-a) \leq -b\log_2 b - (1-b)\log_2 (1-b)$?

Question

If $a,b\in [0,1], a(1-a) \leq b(1-b)$, is it necessary that $-a\log_2 a - (1-a)\log_2 (1-a) \leq -b\log_2 b - (1-b)\log_2 (1-b)$ ?

I tried $a=0.3, b=0.5$ and it worked. Don't see how to prove it in general though.

I can only get down to proving $$-\log_2 a - \log_2(1-a)\leq \log_2b-\log_2(1-b)$$

But I don't see how to get the $a,b$ before the logs in there.

Answer 1

The condition gives $(a-b)(a+b-1)\geq0.$

For $a=b$ we get equality.

Let $a>b$.

The case $0<a<\frac{1}{2}$ is impossible because $a+b-1<0$, which is a contradiction.

Thus, $\frac{1}{2}\leq a<1.$

Now, we can rewrite our inequality in the following form: $$(1-a)^{1-a}a^a\geq(1-b)^{1-b}b^b$$ and since $$\left((1-x)^{1-x}x^x\right)'=(1-x)^{1-x}x^x\ln\frac{x}{1-x}\geq0$$ for all $\frac{1}{2}\leq x<1$, our inequality is proven for $\frac{1}{2}\leq b<1.$

Thus, it's enough to prove our inequality for $0<b<\frac{1}{2}\leq a<1.$

But in this case $0<1-a\leq\frac{1}{2}$ and $1-a\leq b$,

which ends the proof in this case because $(1-x)^{1-x}x^x$ decreases on $\left(0,\frac{1}{2}\right].$

The case $a<b$ is the same.