Prove $(1-x)\ln(1-x)+(1+x)\ln(1+x)\leq 2x^2$ for $0<x<1$

Question

Find the smallest value of $c$ such that $(1-x)\ln(1-x)+(1+x)\ln(1+x)\leq c x^2$ holds for $0<x<1$.

I saw the curve and realized this is true for $c=2$. How can I prove it? What is the smallest $c$ that still makes the inequality valid?

I think we should use the Taylor expansions of the $\ln$s.

Answer 1

To find the smallest value of $c$, use the following method:

Assume $(1-x)\ln (1-x) + (1+x)\ln (1+x) = cx^2$. You have a function in two variables. You can (if it makes it easier for your to visualize), replace $c$ with $y$. Now, you have $y(x)$. Use techniques of calculus to maximize the function.

$$\dfrac{(1-x)\ln (1-x) + (1+x) \ln (1+x)}{x^2} = c$$

With a little calculus, you will find that the derivative has no roots in the reals. So, you need to look at limits.

$$\lim_{x \to 0} \dfrac{(1-x)\ln (1-x) + (1+x) \ln (1+x)}{x^2} = 1$$

$$\lim_{x \to 1} \dfrac{(1-x)\ln (1-x) + (1+x) \ln (1+x)}{x^2} = \ln (4) \approx 1.386$$

That would be our minimum value for $c$.

$$c\ge \ln(4)$$

Answer 2

HINT:

Note that for $t>0$

$$\log(t)\le t-1$$

Answer 3

We have that

• $\ln(1+x)\le x$

• $\ln(1-x)\le -x$

therefore

$$(1-x)\ln(1-x)+(1+x)\ln(1+x)\le -x+x^2+x+x^2= 2x^2$$

Answer 4

Idea to find smallest $c$: Write $f(x):=\dfrac{(1−x)\ln(1−x)+(1+x)\ln(1+x)}{x^2}$ and use calculus to find local maximums or determine if the functions is increasing or decreasing. If the function is increasing or decreasing on $(0,1)$, the smallest value of $c$ can be found by taking a limit at the appropriate endpoint.

Answer 5

Another hint:

If $1>x>0$, one has $\;\ln(1-x)<-x\;$ and $\;\ln(1+x)<x$, and you can make any linear combination of inequalities with positive coefficients.

Answer 6

I flipped the inequalities by mistake. However, I decided to keep this answer as I think it is interesting to know that $$bx^2\leq (1-x)\ln(1-x)+(1+x)\ln(1+x)\leq cx^2\text{ for }x\in(-1,+1)$$ for all $b\leq 1$ and $c\geq \ln(4)$. Other people showed that $c=\ln(4)$ is the smallest value of $c$. I am showing that $b=1$ is the largest value of $b$.

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I claim that $b=1$ works. Observe that $$f(x):=(1-x)\ln(1-x)+(1+x)\ln(1+x)=\int_0^x\,\big(\ln(1+u)-\ln(1-u)\big)\,\text{d}u\,,$$ for all $x\in[0,+1)$. Now, $\ln(1+u)-\ln(1-u)\geq 2u$ for $u\in[0,+1)$; this is because $$\frac{1+u}{1-u}\geq 1+2u+2u^2+2u^3+2u^4\geq 1+2u+2u^2+2u^3+\frac{10}{3}u^4\geq \exp(2u)$$ for all $u\in[0,+1)$. That is, $$f(x)\geq \int_0^x\,(2u)\,\text{d}u=x^2\text{ for }x\in[0,+1)\,.$$ Indeed, we also have $f(x)\geq x^2$ when $x\in (-1,0]$, as $f$ is an even function.

To see why $b=1$ is the largest possible value, you only need to show that $$\lim_{x\to 0}\,\frac{(1-x)\,\ln(1-x)+(1+x)\,\ln(1+x)}{x^2}=1\,.$$ This can be done in many ways, and one method is using L'Hôpital's Rule twice.

Answer 7

The radius of convergence of the Maclaurin series of $(1-x)\log(1-x)+(1+x)\log(1+x)$ is one.
In explicit terms $$\begin{eqnarray*} (1-x)\log(1-x)+(1+x)\log(1+x) &=& x^2+\frac{x^4}{6}+\frac{x^6}{15}+\frac{x^8}{28}+\ldots\\&=&x^2+\sum_{n\geq 2}\frac{x^{2n}}{n(2n-1)} \end{eqnarray*}$$ hence $$ f(x)=\frac{(1-x)\log(1-x)+(1+x)\log(1+x)}{x^2} = 1+\sum_{n\geq 1}\frac{x^{2n}}{(n+1)(2n+1)} $$ is an increasing function on $(0,1)$, ranging from $1$ to $\color{red}{2\log 2}$, which is the optimal $c$-constant.
An improved inequality is $f(x)\leq x^2+(2\log 2-1)x^4$. An improved lower bound is $f(x)\geq \frac{x^2}{1-x^2/6}$.