Logarithm inequality for specific range

Question

I need to show that: $$ \ln(1+x)\left(\ln\left(\frac{1+x}{1-x}\right)+1\right)+\ln(1-x)\ge 0, $$ for $0\le x\le 2/3$. Thanks

Answer 1

Over $[0,1)$ we have $\log\frac{1+x}{1-x}\geq 2x$, hence it is sufficient to prove that over $\left[0,\frac{2}{3}\right]$ we have:

$$ (1+2x)\log(1+x)+\log(1-x) \geq 0 \tag{1}$$ or:

$$ (1-x)\cdot(1+x)^{2x+1}=(1-x^2)\cdot (1+x)^{2x}\geq 1. \tag{2}$$ However, if $x\in\left[\frac{1}{2},\frac{1}{\sqrt{2}}\right]$ we have, by $(1+x)^{\alpha}\geq 1+\alpha x$ for $\alpha\geq 1$: $$ (1-x^2)\cdot (1+x)^{2x} \geq (1-x^2)(1+2x^2) \geq 1, \tag{3}$$ and since over $\left[0,\frac{1}{2}\right]$ we have: $$ (1+x)^{x}\geq 1+\frac{4 x^2}{5}\tag{4} $$ it happens that $(3)$ holds over $\left[0,\frac{1}{2}\right]$, too.

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Update with a proof of $(4)$.

We have that $\log(1+x)$ is a concave function on $J=\left[0,\frac{1}{2}\right]$, hence over $J$: $$ \log(1+x) \geq 2\log\left(\frac{3}{2}\right) x\geq \frac{4x}{5}, $$ then multiplying both sides by $x$ and exponentiating: $$ (1+x)^x \geq \exp\left(\frac{4x^2}{5}\right)\geq 1+\frac{4x^2}{5}.$$