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 -...

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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**