x<(1+x)ln(1+x)<x(1+x)
letf(x)=(1+x)ln(1+x)−x..............(1)
f′(x)=(1+x)(1+x1)+ln(1+x)1−1
f′(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=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)1+x1+ln(1+x)1
=2x+ln(1+x)>0 From this we find that,
g(x)>g(0),whereg(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
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