\(\begin{aligned} & & x+\frac{1}{x} & =y & & |\cdot x\\ & \iff & x^{2}+1 & =xy & & |-1+xy\\ & \iff & x^{2}-xy & =-1 & & |+\left(\frac{y}{2}\right)^{2}\\ & \iff & x^{2}-xy+\left(\frac{y}{2}\right)^{2} & =\left(\frac{y}{2}\right)^{2}-1\\ & \iff & \left(x-\frac{y}{2}\right)^{2} & =\left(\frac{y}{2}\right)^{2}-1 & & |\sqrt{\phantom{\square}}\\ & \iff & x-\frac{y}{2} & =\pm\sqrt{\left(\frac{y}{2}\right)^{2}-1} & & |+\frac{y}{2}\\ & \iff & x & =\frac{y}{2}\pm\sqrt{\left(\frac{y}{2}\right)^{2}-1} \end{aligned}\)
Ob \(x + \frac{1}{x} > y\) oder \(x + \frac{1}{x} < y\) ist, ändert sich also bei
\(x_1 \coloneqq \frac{y}{2}-\sqrt{\left(\frac{y}{2}\right)^{2}-1}\)
und bei
\(x_2 \coloneqq \frac{y}{2}+\sqrt{\left(\frac{y}{2}\right)^{2}-1}\).
N.B. \(x_1\leq x_2\).
\(\begin{aligned} & & x+\frac{1}{x} & >y & & |\cdot x\\ & \stackrel{x>0}{\iff} & x^{2}+1 & >xy & & |-xy\\ & \iff & x^{2}-xy +1 & >0 \end{aligned}\)
Die Parabel der Funktion
\(f(x) = x^{2}-xy +1\)
ist nach oben geöffent. Es gilt also
\(x^{2}-xy+1>0\iff x<x_{1}\vee x>x_{2}\)
und somit
\(x+\frac{1}{x} > y \iff x<x_{1}\vee x>x_{2}\)
