Showing $\lvert \frac{x^2-2x+3}{x^2-4x+3}\rvert\le1\Rightarrow x\le0$ 
$\lvert \frac{x^2-2x+3}{x^2-4x+3}\rvert\le1\Rightarrow x\le0$

How is the proof. If I separate the denominator with triangle inequality,
$\lvert \frac{x^2-2x+3}{x^2-4x+3}\rvert\le \frac{\lvert x^2-2x+3\rvert}{\lvert x^2-2x+3 \rvert-\lvert 2x\rvert}\rvert \ge1$ 
Which means that the triangle inequality should stay a strict inequality, so $x^2-2x+3$ and $2x$ have opposite signs. And the discriminant of the former is negative, therefore the roots are complex, in particular the graph doesn't touch the x-axis. One can insert any value to determine its sign, for example at $0$  gives $3$, so $x$ has to be negative.
Is there another possibility, without much text to prove it ?

WLOG let $$\dfrac{x^2-2x+3}{x^2-4x+3}=\cos y$$ where $y$ is real
$$(1-\cos y)x^2+2x(2\cos y-1)+3(1-\cos y)=0$$
$$\implies x=\dfrac{1-2\cos y\pm\sqrt{(2\cos y-1)^2-3(1-\cos y)^2}}{(1-\cos y)}$$
$$=\dfrac{1-2\cos y\pm\sqrt{\cos^2y+2\cos y-2}}{(1-\cos y)}$$
Now the denominator is non-negative.
Hence we need $$\sqrt{\cos^2y+2\cos y-2}\le2\cos y-1$$
$$\iff\cos^2y+2\cos y-2\le(2\cos y-1)^2\iff3(\cos y-1)^2\ge0$$ which is true
