Supremum of (1/x+1)

Question

THIS IS PART OF MY HOMEWORK, all I need is some guidance:

I am trying to prove that 1 is the supremum of the function $\ f(x)= \frac{1}{(x+1)}$ for every $\ x\gt$ 0. I was able to show that 1 is an upper bound but I can't find a way to show that there's no smaller upper bound than 1 in order to show that it is a supremum.

Answer 1

If $r<1$, $\frac1{x+1}=r\iff x=\frac1r-1$. So, since $\frac1r-1>0$…

Answer 2

if p > 0 is the supremum p < 1, then p != 1/(x+1) (x is arbitrary real number)

so 1 > p > 1/(x+1) > 0

px + p > 1

px > 1 - p > 0

x > (1-p) / p

then for x =< (1-p) / p contradiction

but if supremum is 1, 1 >= 1/(x+1) --> x+1 >= 1 --> x >= 0 --> not contradiction