# inequality x/(x+1)<ln (1+x) , if x+1>0?

### 2 Answers | Add Yours

It has to be determined if the inequality `x/(x+1)<ln (1+x)` holds if x+1>0.

x + 1 > 0

=> x > -1

If the inequality `x/(x+1)<ln (1+x)` holds so does `x/(x+1)-ln (1+x)<0` . But the graph of `y = x/(x+1)-ln (1+x)` is:

At x = 0, `x/(x+1)-ln (1+x)` = 0 - 0 = 0

**The given inequality does not hold, rather it should be ` x/(x+1)<=ln (1+x)` **

The request of the problem is vague, hence, supposing that you need to test if the inequality holds, you should move all members of inequality to one side and you need to check if the new function is negative for al x > -1, such that:

`f(x) = x/(x + 1) - ln(1 + x) < 0`

You need to use the theory of derivative of the function such that:

`f'(x) < 0 => f(x) < 0`

Hence, you need to test if derivative of the function is negative, such that:

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

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

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

You need to bring the fractions to a common denominator, such that:

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

Since for all `x>-1` , the derivative `f'(x) = -x/((x+1)^2)` rests negative, yields that `f(x) < 0.`

**Hence, since `f(x) = x/(x + 1) - ln(1 + x) < 0` , then the inequality `x/(x + 1) < ln(1 + x)` holds for `x > -1.` **

### Hide Replies ▲

The statement `f'(x) < 0 => f(x) < 0` is not true.

For example consider f(x) = sin (x) - x

f'(x) = cos x - 1

It is seen in the graph above that for values of x in `{-oo, 0}` though `f'(x) < 0` , `f(x) > 0`