Question

Using LMV, prove that log(1+x) > 2x/(1+x) , if x>0

Answer 1

Solution:

Let $f(x)=\log (1+x)$  in [0, x]

Since f(x) satisfies the condition of L.M.V. theorem in [0, x] , there exists $\theta(0<\theta<1)$  such that

$\begin{aligned} &\frac{f(x)-f(0)}{x-0}=f^{\prime}(\theta x) \\ &\Rightarrow \quad \frac{\log (1+x)}{x}=\frac{1}{1+\theta x} \end{aligned}$

Now, $\quad 0<\theta<1, x>0 \Rightarrow \theta x<x$

 $\Rightarrow \quad 1+\theta x<1+x \ \Rightarrow \quad \frac{1}{1+\theta x}>\frac{1}{1+x} \\\Rightarrow \quad \frac{x}{1+\theta x}>\frac{x}{1+x} \ ...(i)$

Again $0<\theta<1, x>0$

$\begin{array}{ll} \Rightarrow & \theta x>0 \\ \Rightarrow & 1+\theta x>1 \\ \Rightarrow & \frac{1}{1+\theta x}<1 \\ \Rightarrow & \frac{x}{1+\theta x}<x\ ...(ii) \end{array}$

From (i) and (ii), we get,

$\frac{x}{1+x}<\log (1+x)<x$