Knowing f(x)=lnx-[2(x-1)/(x+1)] demonstrate that for x>1 lnx>[2(x-1)/(x+1)

We've noticed that if we bring the paranthesis
2(x-1)/(x+1) on the left side of the inequality, which has to be demonstrated, is no one
else but the function f(x) itself.

So, we have to
demonstrate that f(x)>0. For proofing this, we have to verify if the first
derivative of the function is also
positive.

f'(x)=[lnx-2(x-1)/(x+1)]'=

=(1/x)-{[2(x-1)'*(x+1)-2(x-1)*(x+1)']/(x+1)^2}=

=(1/x)-(2x+2-2x+2)/(x+1)^2=(1/x)-(4)/(x+1)^2=

=[(x+1)^2-4x]/x*(x+1)^2=(x^2+2x+1-4x)/x*(x+1)^2=

=(x^2-2x+1)/x*(x+1)^2=(x-1)^2/x*(x+1)^2

But
for x>1, (x-1)^2>0 and x*(x+1)^2>0, so
f(x)>0