Question

How do we show that for all x>0 , that x + ln(1+x) > (x^2/(2(1+x))) ?

I differentiated the function w.r.t x and got f'(x) =((x+2)^2)/(2(1+x)) , and for x>0, f'(x)>0.
Does this prove what's required?

Answer 1

x + ln(1+x) > (x^2/(2(1+x)))

or ln(1+x) > x^2/(2+2x) - x
ln(1+x) > (x^2 - 2x - 2x^2)/(2+2x)
ln(1+x) > (-x^2 - 2x)/(2+2x)
ln(x+1) > -x(x+2)/(2x+2)

now, we know ln(1) = 0 , and since x is positive ln(1+x) must be positive.
Looking at the right side we see that for x> 0, both x+2 and 2x+2 must be positive and a positive divided by a positive must be positive.
but that is multiplied by -x, clearly a negative
so the right side of the above inequation is negative
So we have "a positive > a negative" which of course is true.

test it on your calculator, eg let x = .0001 and the x= 5000