Why $\frac{(1-x)\log(1-x)}{x\log x}$ is increasing on (0, 1)?

Question

Let $f(x)=\frac{(1-x)\log(1-x)}{x\log x}$. I find one claims this function is increasing on $(0,1)$. While I find the numerator of $f'(x)$ is $$-x\log x -\log x\log(1-x)-(1-x)\log(1-x).$$ I am wondering why this is non-negative?

Answer 1

Consider the function $g(x)=\frac{\log(1-x)}{x}$. Then

$$f(x)=\frac{g(x)}{g(1-x)}$$

Take a derivative of $g$:

$$g'(x)=-\frac{x+(1-x)\log(1-x)}{x^2(1-x)}$$

Consider the auxiliary function $h(x)=x+(1-x)\log(1-x)$. We use a standard argument to show that it is positive. It's derivative is $h'(x)=-\log(1-x)$ which is positive in the interval $(0,1)$ and that means that h is increasing. However $h(0)=0~,~h(1)=1$ which proves given it's monotonicity, that $h(x)\geq 0, x\in(0,1)$.

Therefore we finally conclude that in the interval $(0,1)$ $g'(x)\leq 0$ and thus g is decreasing. Knowing this, we can also conclude that $g(1-x)$ is increasing in the same interval since the transformation $x\to 1-x$ reverses monotonicity but leaves the interval invariant. Finally we see that $1/g(1-x)$ is decreasing since $g(1-x)$ is non-vanishing in this interval and therefore it has the same sign (in fact, it is negative).

Now note that the functions $u_1(x)=-g(x)$ and $u_2(x)=-1/g(1-x)$ are both increasing and positive. One can easily show that multiplying increasing, positive functions results in an increasing, positive function as well. Since $f(x)=u_1(x)u_2(x)$ we conclude that $f$ is increasing and this concludes the proof.

Answer 2

Using taylor series: $$\frac {\ln(1-x)}{x}=-\frac {x+\frac {x^2}2+...}x=- (1+\frac x2+...)$$ Which is decreasing, similarly: $$\frac {\ln(x)}{1-x}=-\frac{\ln(1-(1-x))}{1-x}=-(1+\frac{1-x}2+...)$$ Which is increasing, then its reciprocal $\frac {1-x}{\ln(x)}$ is decreasing.

Since these two are negative, then they multiplied together is increasing.

Answer 3

Using the argument from $\frac{\log(x)}{1-x}$ is increasing (and similar to DinosaurEgg's solution):

$$ h(x) = -\frac{\log(x)}{1-x} = \frac{1}{x-1} \int_1^x \frac{dt}{t} = \int_0^1 \frac{ds}{1+s(x-1)} $$ is positive and decreasing for $x \in (0, 1)$. Then $$ h(1-x) = -\frac{\log(1-x)}{x} $$ is positive and increasing in that interval, so that $$ f(x)=\frac{(1-x)\log(1-x)}{x\log x} = \frac{h(1-x)}{h(x)} $$ is increasing.

Answer 4

(Not a complete answer but hopefully close enough)

You want to show that $$ x \log(x)+ (1-x)\log(1-x) \le - \log(x) \log(1-x)$$ Divide on both sides by $\log(x) \log(1-x)$ and this is $$ \frac{x}{\log(1-x)} + \frac{1-x}{\log(x)} \le -1 $$

This is of the form $$ f(x) + f(1-x) \le c$$ where $f(x) = \frac{x}{\log(1-x)}$. I don't know if there is a nice way to show this.

But the not-so-nice way is to Taylor expand $$ \frac{x}{\log(1-x)} = -1 + \frac{x}{2} + \frac{x^2}{12} + \frac{x^3}{24} + \dotsm $$ $$ \frac{1-x}{\log(x)} = -1 + \frac{1-x}{2} + \frac{(1-x)^2}{12} + \frac{(1-x)^3}{24} + \dotsm $$

Now you add them up, you see that the RHS becomes $$ -2 + \frac{1}{2} + \frac{x^2 + (1-x)^2}{12} + \frac{x^3 + (1-x)^3}{24} + \dotsm $$ $$ \le -2 + 1 + \frac{1}{12} + \frac{1}{24} + \dotsm $$

But then showing that the last sum is $\le -1$, seems clear to me but maybe you might want more concrete justification...