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

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### 2 Answers

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**

f(x) = lnx - [2(x-1)/(x+1)] . To demonstrate that forx>1, lnx > [2(x-1)/(x+1).

f(1) = ln1 - 1 = 2(1-1)/(1+1) = 0.

We examine weather f(x) is increasing function for which f'(x) >0

f'(x) = (lnx)' - {2(x-1)/x+1)}'

f'(x) = 1/x - {2 (x+1)' - 2(x-1)(1)'}/(x+1)^2

f'(x) = 1/x - (2x+2 -2x+2)/(x+1)^2

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

f'(x) = {(x+1)^2 -4}/(x+1)^2 > o for x>1.

This estblishes that f(x) = 0 at x= 1 and f(x) is an increasing function.

Therefore f(x) i s >0 for x>1. This implies lnx-2(x-1)/(x+1) > 0

Or lnx > 2(x-1)/(x+1).