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 ?

What you want is
$$\left|\frac{x^2-2x+3}{x^2-4x+3}\right|\le1\iff-1\le\frac{x^2-2x+3}{x^2-4x+3}\le1$$
Beginning with the left inequality:
$$-1\le\frac{x^2-2x+3}{x^2-4x+3}\iff\frac{x^2-2x+3}{x^2-4x+3}+1\ge0\iff\frac{2x^2-6x+6}{x^2-4x+3}\ge0\iff$$
$$\frac{x^2-3x+3}{(x-1)(x-3)}\ge 0\iff (x-1)(x-3)>0\;\text{ (why?)}\implies x<1\;\text{or}\;x>3$$
Now you do the other inequality and take the intersection of both solution sets.
