Proving: $\frac{2x}{2+x}\le \ln (1+x)\le \frac{x}{2}\frac{2+x}{1+x}, \quad x>0.$

Question

$$\begin{equation}\frac{2x}{2+x}\le \ln (1+x)\le \frac{x}{2}\frac{2+x}{1+x}, \quad x>0\end{equation}$$

I found this inequality in this paper: http://ajmaa.org/RGMIA/papers/v7n2/pade.pdf (Equation (3)).

How exactly can I prove it? I tried induction but to no avail...

Answer 1

Let $f(x) = \frac{2x}{x+2}$, $g(x) = \ln(1+x)$, $h(x) = \frac{x^2 + 2x}{2x+2}$.

$f(0) = 0$, $g(0) = 0$, $h(0) = 0$.

$f' = \frac{2(x+2)-2x}{(x+2)^2} = \frac{4}{(x+2)^2}$, $g' = \frac{1}{1+x}$, $h'= \frac{(2x+2)(2x+2)-2(x^2+2x)}{(2x+2)^2}=\frac{2x^2+4x+4}{(2x+2)^2}$

Now it is easy to check that $$f'=\frac{4}{(x+2)^2} \leq g' = \frac{1}{1+x} \leq h'=\frac{2x^2+4x+4}{(2x+2)^2}$$

Answer 2

Hint :

Consider $$f(x)=\ln (1+x)-\frac{2x}{2+x}$$and $$g(x)=\frac{x(x+2)}{2(x+1)}-\ln(1+x)$$Show $f$ and $g$ are monotone increasing functions..

Answer 3

I have a shorter proof: by the AM-GM inequality, $$\forall t>0,\qquad \frac{1}{1+t}\geq \frac{1}{\left(1+\frac{t}{2}\right)^2}\tag{1} $$ as well as: $$\forall t>0,\qquad \frac{1}{1+t}\leq \frac{1+(1+t)^2}{2(1+t)^2}.\tag{2}$$ Now it is enough to integrate both sides of $(1)$ and $(2)$ over $(0,x)$ to prove the statement.