Answer to Question #316317 in Calculus for Bless

Question

Prove that x(1 + x) > (1 + x) In(1 + x) > x

Answer 1

$x<(1+x)ln(1+x)<x(1+x)$

$let f(x)=(1+x)ln(1+x)-x..............(1)$

$f'(x)=(1+x)(\frac{1}{1+x})+ln(1+x)1-1$

$$f'(x)=1+ln(1+x)-1=ln(1+x)>0$$

i.e. $x>0; f(x)>f(0), where f(0)=(1+0)ln(1+x)-0=0$

Hence, f(x)>0

We use 1 above

$(1+x)ln(1+x)-x>0 ;$

$$(1+x)ln(1+x)>x.....(M)$$

Let g(x)=x(1+x)-(1+x)ln(1+x)........(2)

$g'(x)=(2x+1)-((1+x)\frac{1}{1+x}+ln(1+x)1$

$$=2x+ln(1+x)>0$$

From this we find that,

$g(x)>g(0), where \,g(0)=0(1+0)-(1+0)ln(1+0)=0$

g(x)>0...use (2)

$x(1+x)-(1+x)ln(1+x)>0$

$x(1+x)>(1+x)ln(1+x).........(N)$

From (N)and (M)

x<(1+x)ln(1+x)<x(1+x)

Hence it's clear that,

$$x(1+x)>(1+x)ln(1+x)>x$$