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

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 $f(x)\geq \frac{x^2}{1-x^2/6}$.
