Inequality with $\ln$ and powers

Question

I have a long inequality which I reduced to $$\left(1-(1-x)^t\right)\left(1-\frac{x}{2}\right)^t\ln(1-\frac{x}{2}) < \left(1-(1-\frac{x}{2})^t\right)\left(1-x\right)^t\ln(1-x),$$ where $t\geq 1$ and $x\in(0,1).$ Any hints on proving it.

Numerical simulations show that the inequality holds.

Answer 1

Using MPW's comment, I rewrote the inequality as

$$\frac{\left(1-\frac{x}{2}\right)^t}{\left(1-(1-\frac{x}{2})^t\right)}\ln(1-\frac{x}{2}) < \frac{\left(1-x\right)^t}{\left(1-(1-x)^t\right)}\ln(1-x),$$

so both either side can be written as $\frac{y^t}{1 - y^t}\ln y$. Differentiate the function w.r.t. $y$ to obtain $$\frac{ty^{t-1} + ty^{t-1}y^t}{(1-y^t)^2}\ln y + \frac{y^{t-1}}{1-y^t},$$ which is negative only if

\begin{equation} \frac{1-y^t}{t(1+y^t)} <- \ln y.\tag{*} \end{equation}

Denote the LHS term by $f_1(y)$ and the RHS by $f_2(y)$. Both sides of the inequality are positive. We have: $$\lim_{y\to 0} f_1(y) = \frac{1}{t},\ \ \ \lim_{y\to 0} f_2(y) = \infty,$$ $$\lim_{y\to 1} f_1(y) = 0,\ \ \ \lim_{y\to 1} f_2(y) = 0,$$

and $f'_1(y) = -\frac{2ty^t}{y(1+y^t)^2}<0$ and $f'_2(y) = -\frac{1}{y}<0$. Finally, $f'_1(y) - f'_2(y) = \frac{1 + 2y^t(t-1)+y^{2t}}{y(1+y^t)^2}>0$. This shows the inequality (*) holds, or that $\frac{y^t}{1 - y^t}\ln y$ is decreasing in $y$. This means that the LHS of the original inequality is helds as $1-\frac{x}{2} > 1-x$.

Answer 2

For $\,x>0\,$ we have $\,\displaystyle e^x>\frac{e^x-1}{x}\,$ (using the series of $\,e^x\,$) .

It follows $\,\displaystyle z\ln z >z-1\,$ for $\,z:=e^x>1\,$ .

Let $\,\displaystyle f(z):=\frac{\ln z}{1-z}\,$ with $\,z>1\,$ .

We get $\,\displaystyle f’(z)=\frac{1-z+z\ln z}{z(1-z)^2}>0\enspace$ which means that $\,f(z)\,$ is increasing.

Let $\,\displaystyle z:=\frac{1}{(1-\frac{x}{a})^t}>1\,$ with $\,a,t\geq 1\,$ and $\,0<x<1\,$ .

Then $\,z\,$ decreases for increasing $\,a\,$ and any fixed $\,x\in (0,1)\,$ and it means that $\,\displaystyle f\left(\frac{1}{(1-\frac{x}{a})^t}\right)\,$ decreases too.

It e.g. (see the question) follows: $\enspace\displaystyle \frac{1}{t} f\left(\frac{1}{(1-\frac{x}{2})^t}\right) < \frac{1}{t}f\left(\frac{1}{(1-x)^t}\right)$

Answer 3

Let $u=(1-x)^t$ and $v=(1-x/2)^t$. Note that, for $t\gt1$ and $0\lt x\lt1$, we have

$$0\lt u\lt v\lt 1$$

Now, on multiplying both sides of the OP's inequality by $t$, and bringing that $t$ into the logarithms, we have the equivalent inequality

$$(1-u)v\ln v\lt(1-v)u\ln u$$

which can be proved by showing that

$$f(u)={u\over1-u}\ln u$$

is a strictly decreasing function for $u\in(0,1)$. Doing so is a routine matter of showing that

$$f'(u)={1-u+\ln u\over(1+u)^2}$$

and $(1-u+\ln u)'=-1+{1\over u}\gt0$ for $0\lt u\lt1$, so that $f'(u)\lt f'(1)=0$.