Question

Show that `ln(1+x) < x-(x^2)/(2(1+x)), AA x>0`?

Answer 1

See explanation below

Explanation:

Simplify the expression at the second member:

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

We can further simplify adding and subtracting `1` to the numerator:

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

Consider now:

`(df)/dx = 1/2+1/2 1/(1+x)^2`

`d/dx ln(1+x) = 1/(1+x)`

We have:

`f(x)-ln(1+x) = f(0)+ln(1+0)+ int_0^x g(t)dt = 1/2-1/2+0+int_0^x g(t)dt = int_0^x g(t)dt`

where `g(x) = (df)/dx -d/dx ln(1+x)`

so:

`g(x) = 1/2 +1/2 1/(1+x)^2 -1/(1+x)`

`g(x) =((1+x)^2+1+2(1+x))/(2(1+x)^2)`

`g(x) =(1+2x +x^2+1+2+2x)/(2(1+x)^2)`

`g(x) = (x^2+4x+4)/(2(1+x)^2)`

`g(x) = (x+2)^2/(2(1+x)^2)`

Clearly `g(x) >0` for every `x` and then for `x>0`:

`int_0^x g(t)dt > 0`

so:

`f(x)-ln(1+x) > 0`

`f(x) > ln(1+x)`

Answer 2

See below.

Explanation:

Calling `f(x)=x-log_e(x+1)+x^2/(2(x+1))` we have

`f(0) = 0`

To demonstrate that `f(x) gt 0` for `x > 0`

we will show that `f(0)` is a minimum point. So we will determine the stationary points.

`(df)/(dx) = (4 x + 3 x^2)/(2 (1 + x)^2) = 0`

solving we get at `x = -4/3` and `x = 0`

qualifying the stationary points we get

`(d^2f)/(dx^2) = (x+2)/(x+1)^3` assuming the values

`((x,(d^2f)/(dx^2),"type" ),(-4/3,-18,"maximum"),(0,2,"minimum"))`

so concluding, `x=0` is a minimum of `f(x)` so

for `x > 0` we have `f(x) > 0` and then

`log_e(x+1) < x+ x^2/(2 (x + 1))`

Answer 3

See the Proof given in the Explanation.

Explanation:

Define fun. `f`, by, `f(x)=x-x^2/(2(1+x))-ln(1+x), x >0.`

We have, `x^2/(1+x)={(x^2-1)+1}/(1+x)=(x^2-1)/(1+x)+1/(1+x)=x-1+1/(1+x).`

`:. f(x)=x-1/2{x-1+1/(1+x)}-ln(1+x),`

`=x/2+1/2-1/(2(1+x))-ln(1+x), i.e.,`

`f(x)=(x+1)/2-1/(2(1+x))-ln(1+x), x>0.`

`:. f'(x)=1/2-1/2{-1/(1+x)^2}-1/(1+x),`

`=1/2+1/(2(1+x)^2)-1/(1+x),`

`={(1+x)^2+1-2(1+x)}/(2(1+x)^2),`

`rArr f'(x)=x^2/(2(1+x)^2) (x>0).`

Clearly, `f'(x)>0,` which means that, `f" is "uarr AA x >0.`

`:. x > 0 rArr f(x) > f(0)=0, i.e., f(x) > 0.`

`rArr x-x^2/(2(1+x))-ln(1+x) > 0, AA x > 0" or, equivalently,"`

`ln(1+x) < x-x^2/(2(1+x)), x > 0.`

#Enjoy Maths.!