We have to prove that [(x+1)*lnx]/2 > x-1

[(x+1)*lnx]/2 > x-1

=> [(x+1)*lnx] > 2(x-1)

=> ln x > 2(x-1)/(x+1)

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

If we find the derivative of ln x - 2(x-1)/(x+1)

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

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

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

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

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

We see that (x-1)^2 is always positive, also x*(x+1)^2 is always positive for positive values of x for which ln x exists.

So (x-1)^2/x*(x+1)^2 > 0

**Therefore we have proved that [(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**