Lower bound for the function $h(x)=(1+1/x)\log(1+x)-1$ for $x > 0$.

Question

I am reading a paper and in the paper, there is a result having no proof. I was wondering if anyone could give me a hint on this result. The result is the following: $$ h(x)\geq \begin{cases} x/4, & \text{if } x\leq4;\\ (1/2)\log x, & \text{if } x\geq4, \end{cases} $$ where the function $h$ is defined for all $x > 0$ by $$h(x)=(1+1/x)\log(1+x)-1.$$

Thanks in advance.

Answer 1

For $0<x\leq 4$ we need to prove that $$\left(1+\frac{1}{x}\right)\ln(1+x)-1\geq\frac{x}{4}$$ or $f(x)\geq0,$ where $$f(x)=\ln(1+x)-\frac{4x+x^2}{4(1+x)}.$$ But, $$f'(x)=\frac{x(2-x)}{3(1+x)^2},$$ which says that $$\min_{0\leq x\leq4}f=\min\{f(0),f(4)\}=f(0)=0.$$ By the same way you can prove the second inequality:

after calculating of derivative use AM-GM.