How to verify the inequality: {(x+1)*lnx}/2>x-1

First, we'll try to re-write the expression in an easier
manner:

(x+1)*lnx >
2(x-1)

Now, we'll divide both sides by
(x+1):

ln x >
2(x-1)/ (x+1)

We'll subtract the ratio 2(x-1)/ (x+1) both
sides:

ln x - 2(x-1)/ (x+1) >
0

We'll note the given
expression:

f(x) = lnx -
2(x-1)/(x+1)

We'll have to demonstrate that
f(x)>0.

To prove that a function is increasing,
we'll have to prove that the first derivative of the function is
positive.

We'll calculate the first
derivative:

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

The ratio from the expression of the
function will be differentiated using the quotient
rule.

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

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

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

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

We notice that for any real value of x,
(x-1)^2>0 and x*(x+1)^2>0, so
f(x)>0.

f(x) =
lnx-2(x-1)/(x+1)

lnx-2(x-1)/(x+1) > 0
q.e.d