In GMB 5/2005 magazine, pages 234 (Romanian version) and 236 (English version).
The problem says :
C:2870. Find the minimum value of : $E(x)=\frac{x^2-2x-1}{x^2-2x+3}$, for $x\in\mathbb{R}$.
[Author :] Vasile PREDAN, Curtea de Argeș
ANSWER CiP
$-1=E(1)\leqslant E(x)<1$
Solution CiP We will imitate the solution from the Post here.
The number $\lambda$ is a value of $E(x)\;\;\Leftrightarrow$
$\Leftrightarrow (\exists)x\in\mathbb{R}\;\;\mathbf{s.t.}\;\;E(x)=\lambda$
$\Leftrightarrow (\exists)x\in\mathbb{R}\;\;\mathbf{s.t.}\;\;x^2-2x-1=\lambda \cdot (x^2-2x+3)$
$\Leftrightarrow (\exists)x\in\mathbb{R}\;\;\mathbf{s.t.}\;\;(1-\lambda)x^2+2(\lambda-1)x-(3\lambda+1)=0 \tag{1}$
$\Leftrightarrow$ the quadratic equation (1) has solution(s).
[For $\lambda=1$ equation (1) isn't quadratic both then in hasn't solution.].
Using the half discriminant ($\Delta'=(b')^2-ac$) formula, the roots of equation (1) are
$x_{1,2}=\frac{1-\lambda\pm\sqrt{2-2\lambda^2}}{1-\lambda} \tag{2}$
The requirement that the roots be real numbers, $|\Delta'\geqslant 0$ implies $2-2\lambda^2\geqslant 0\;\Leftrightarrow$
$\Leftrightarrow\;\lambda^2\leqslant 1\;\Leftrightarrow\;-1\leqslant \lambda<1.$ From (2) we obtain the expressions of the roots as follows $x_{1,2}=1\pm \sqrt{\frac{2(1+\lambda)}{1-\lambda}}.$
$\blacksquare$
Writing differently, $E(x)=1-\frac{4}{(x-1)^2+2}\;\;\Rightarrow\;\; -1=1-\frac{4}{0+2}\leqslant 1-\frac{4}{(x-1)^2+2}<1$, with the same answer.
