Nice Exercise asked by Kunny Shogun in many math group and its solution base on MVT and Continuous functions

Exercise:

Let $a,b\in {{\mathbb{R}}^{+}}$ such that $a\ne b$ , show that, ${{\left( \frac{1+a}{1+b} \right)}^{\frac{1}{a-b}}}<e$  *

Solution: Let $f\left( x \right)=\ln \left( 1+x \right)$ and since $f$ is continuous on $\left[ a,b \right]$

And differentiable on $\left( a,b \right)$ then by MVT

$\exists c\in {{\mathbb{R}}^{+}}$ Between $a$ and $b$ such that $f'\left( c \right)=\frac{f\left( a \right)-f\left( b \right)}{a-b}$

As $a,b,c>0$ so $f'\left( x \right)=\frac{1}{1+x}\Leftrightarrow f'\left( c \right)=\frac{1}{1+c}<1\Leftrightarrow \frac{f\left( a \right)-f\left( b \right)}{a-b}<1$

Hence $\frac{\ln \left( 1+a \right)-\ln \left( 1+b \right)}{a-b}<1\Leftrightarrow \ln \left( 1+a \right)-\ln \left( 1+b \right)<a-b\Leftrightarrow \ln \frac{1+a}{1+b}<a-b$

But $e$ is continuous and increasing function

So ${{e}^{\ln \left( \frac{1+a}{1+b} \right)}}<{{e}^{a-b}}\Leftrightarrow \sqrt[a-b]{{{e}^{\ln \left( \frac{1+a}{1+b} \right)}}}<\sqrt[a-b]{{{e}^{a-b}}}\Leftrightarrow {{\left( \frac{1+a}{1+b} \right)}^{\frac{1}{a-b}}}<e$ 

*___________________________________________________
  -2015 Hitotsubashi University entrance exam /Economic ( 2nd Exam )