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

Question

For all $-1<x<1$, prove: $$(1+x)\ln(1+x) + (1-x)\ln(1-x) \leq 2x^2$$

I am sure we should use the Jensen's inequality or the Taylor expansion of $\ln(1+x)$ and $\ln(1-x)$; however, I was not able to do that so far. Any idea?

Answer 1

Hint
If you want to use Jensen, note $\ln(1+t)$ is concave, so $$(1+x)\ln(1+x)+(1-x)\ln(1-x) \leqslant 2\ln(1+x^2) \leqslant 2x^2$$

where the last step is from $\ln(1+t)\leqslant t$, another consequence of concavity.

Answer 2

We can known that $\ln(1+x)\le x$ is True when $x > -1 $

The polit of $y=x$ and $y=\ln(1+x)$

Then$\ln(1-x)\le -x$ when $-x > -1 \Rightarrow x < 1$

So in $-1 > x > 1 $, we can conclude that

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

Prove done, good luck to you