If $|x| \le 2$, prove that $\frac{|x^2-4x+3|}{|x^2-2x+2|} \le 15$.

Question

If $|x| \le 2$, prove that $\frac{|x^2-4x+3|}{|x^2-2x+2|} \le 15$.

I am stuck at the step $|x-3|/|x-1|$ and not able to continue it

Answer 1

Note that $$|x|\le 2 \implies \frac{|x^2-4x+3|}{|x^2-2x+2|} \le \frac{|x^2|+|4x|+3}{(x-1)^2+1} \le \frac{4+8+3}{1} =15$$

Thus if $|x|\le 2$, then $\frac{|x^2-4x+3|}{|x^2-2x+2|}\le 15$

Answer 2

As a traditional alternative to the clever and short solution by Mohammad Riazi-Kermani, note that $x^2-2x+2>0$ then

$$\frac{|x^2-4x+3|}{|x^2-2x+2|} \le 15\implies\frac{|(x-3)(x-1)|}{x^2-2x+2} \le 15$$

then consider two cases with also $|x|\le 2$

• $x<1 \quad \lor \quad x>3 \implies -2\le x<1 \implies \frac{(x-3)(x-1)}{x^2-2x+2} \le 15$
• $1\le x\le3\implies 1\le x\le2 \implies \frac{(x-3)(x-1)}{x^2-2x+2} \ge 15$

Answer 3

Let $$ y =|\frac{x^2-4x+3}{x^2-2x+2}|$$ $ y =\frac{x^2-4x+3}{x^2-2x+2}$ or $y = -\frac{x^2-4x+3}{x^2-2x+2}$

First let us take the former case in which the expression is positive

$$ y =\frac{x^2-4x+3}{x^2-2x+2} $$ $$ y(x^2-2x+2) =x^2-4x+3$$ $$(y-1)x^2-(2y-4)x +(2y-3)=0$$

Now we know that $|x|\le2$. This means that $x \in [-2,2]$. This shows that x is a real number and will have real values.

In the expression above, x will only assume real values if the determinant is greater than or equal to zero.

$$(2y-4)^2-4(2y-3)(y-1) \ge 0$$ I hope you can take it from here. Just solve the inequality above and check the second case. Let me know if you don't find the answer

Answer 4

Yet another way:

$$\left| \dfrac{x^2-4x+3}{x^2-2x+2} \right| =\left|1- \frac{2x-1}{x^2-2x+2}\right| \leqslant 1+\left|\frac{2x-1}{(x-1)^2+1} \right| \leqslant 1+\left|\frac51 \right| <15$$ for $x \in [-2, 2]$