Question #316317

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

1
Expert's answer
2022-03-24T03:12:03-0400

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

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

f(x)=(1+x)(11+x)+ln(1+x)11f'(x)=(1+x)(\frac{1}{1+x})+ln(1+x)1-1


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


i.e. x>0;f(x)>f(0),wheref(0)=(1+0)ln(1+x)0=0x>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>0 ;


(1+x)ln(1+x)>x.....(M)(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)11+x+ln(1+x)1g'(x)=(2x+1)-((1+x)\frac{1}{1+x}+ln(1+x)1


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

From this we find that,

g(x)>g(0),whereg(0)=0(1+0)(1+0)ln(1+0)=0g(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)>0x(1+x)-(1+x)ln(1+x)>0

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


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