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

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 ?

  • $\begingroup$ Consider two cases: $x^2-2x+3\leq x^2-4x+3$, or $x^2-2x+3\leq -(x^2-4x+3)$. You will find that you only have the first case, that is, $x^2-2x+3\leq x^2-4x+3$, which implies that $x\leq0$. $\endgroup$ – m-agag2016 Dec 19 '16 at 11:16

What you want is


Beginning with the left inequality:


$$\frac{x^2-3x+3}{(x-1)(x-3)}\ge 0\iff (x-1)(x-3)>0\;\text{ (why?)}\implies x<1\;\text{or}\;x>3$$

Now you do the other inequality and take the intersection of both solution sets.

  • $\begingroup$ $(x-1)(x-3)>0$ in line 4 $\endgroup$ – Khosrotash Dec 19 '16 at 11:41
  • $\begingroup$ @Khosrotash Excellent catch! Thanks, edited. $\endgroup$ – DonAntonio Dec 19 '16 at 11:51

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

  • $\begingroup$ @DonAntonio, I meant this. Thanks $\endgroup$ – lab bhattacharjee Dec 19 '16 at 12:02

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