Let $S=\{\frac{x+1}{x^2+1}:x\in \mathbb{Q},x>0\}$. Only use the knowledge about supremum and infimum. Prove that $\sup S=\frac{1+\sqrt{2}}{2}$.
I know the basic method the find the supremum and infimum, e.g. Show that $\frac{1+\sqrt{2}}{2}$ is an upper bound for S and assume that the supremum is less than it and find a contradiction. Alternatively, show that the supremum $u\le\frac{1+\sqrt{2}}{2}$ and $u\ge\frac{1+\sqrt{2}}{2}$. However, I cannot find an explicit easy way to derive it. Can someone help me?
Let $x$ be a positive rational number. We compute: $$ \begin{aligned} S-\frac{x+1}{x^2+1} &=\frac {1+\sqrt 2}2-\frac{x+1}{x^2+1} =\frac {1+\sqrt 2}{2(x^2+1)}\left(x^2 + 1-\frac 2{\sqrt 2+1}(x+1)\right) \\ &=\frac {1+\sqrt 2}{2(x^2+1)}\left(x^2 + 1-2(\sqrt 2-1)(x+1)\right) \\ &=\frac {1+\sqrt 2}{2(x^2+1)}\ \Big(\ x^2 -2(\sqrt 2-1)x + (\sqrt 2-1)^2\ \Big) \\ &=\frac {1+\sqrt 2}{2(x^2+1)}\ \Big(\ x - (\sqrt 2-1)\ \Big)^2 \\ &\ge 0\ . \end{aligned} $$ This implies that $S$ is an upper bound for the set of values of the expression $(x+1)/(x^2+1)$, when $x$ runs in the set of positive rational numbers.
On the other side, the real number $(\sqrt 2-1)$ can be approximated by rational numbers to each wanted precision. (We know $\sqrt 2$ from somewhere. We use here, that it is approximable by rational numbers. Either take the decimal representation and truncate it, or use the continued fraction for $\sqrt 2$.) Pick an $\epsilon>0$, and it is enough to take one with $\epsilon<\frac 23$. Then there is an $r>0$, $r$ rational, with $|r-(\sqrt 2-1)|<\epsilon$. From the above it is easy to estimate: $$ \left|S-\frac{r+1}{r^2+1}\right| = \frac {1+\sqrt 2}{2(r^2+1)}\ \Big|\ r - (\sqrt 2-1)\ \Big|^2 < \frac {1+\sqrt 2}{2(0^2+1)}\ \epsilon^2 <\frac 32\epsilon^2<\epsilon \ . $$ So we get as close enought to $S$ as we want...