1. ## Absolute value inequality (and a square root, sadly)

Hi guys,

Given: $\displaystyle \sqrt{|1-x|} \geq x$
Expanding absolute value: $\displaystyle x > 1 \wedge {\color{Red} \sqrt{x-1} \geq x} \vee x \leq 1 \wedge \sqrt{1-x} \geq x$
Note: the inequality marked in red is never true. I thinks this OK because the given is also never true for x > 1.

Questions: Have I expanded the absolute value sign correctly? If not: any hints?

Many thanks

2. ## Re: Absolute value inequality (and a square root, sadly)

Hi edg85!

Originally Posted by edg85
Hi guys,

Given: $\displaystyle \sqrt{|1-x|} \geq x$
Expanding absolute value: $\displaystyle x > 1 \wedge {\color{Red} \sqrt{x-1} \geq x} \vee x \leq 1 \wedge \sqrt{1-x} \geq x$
Note: the inequality marked in red is never true. I thinks this OK because the given is also never true for x > 1.

Questions: Have I expanded the absolute value sign correctly? If not: any hints?

Many thanks
Yes.
All is well.

3. ## Re: Absolute value inequality (and a square root, sadly)

Originally Posted by edg85
Given: $\displaystyle \sqrt{|1-x|} \geq x$

STOP, step back and think. This is a beautiful thought question.

Is it true for any $\displaystyle x\le 0~?$

For what $\displaystyle x>0$ is it true?

When is $\displaystyle x>0\text{ and }|x-1|>x^2\text{ true ?}$

Did you plot $\displaystyle \sqrt{|1-x|}-x~?$

4. ## Re: Absolute value inequality (and a square root, sadly)

Hi, thanks for the replies. @Plato yes i have a plot, handdrawn actually. Something wrong? Have i expanded the absolute value correctly in my first post? If so i will post the rest of my solution asap.

5. ## Re: Absolute value inequality (and a square root, sadly)

Originally Posted by edg85
Hi, thanks for the replies. @Plato yes i have a plot, handdrawn actually. Something wrong? Have i expanded the absolute value correctly in my first post? If so i will post the rest of my solution asap.
Can you show why
$\displaystyle x\le \frac{-1+\sqrt5}{2}$ is a solution?

6. ## Re: Absolute value inequality (and a square root, sadly)

Originally Posted by Plato
Can you show why
$\displaystyle x\le \frac{-1+\sqrt5}{2}$ is a solution?
Sure! Let's go:

Given: $\displaystyle \sqrt{|1-x|} \geq x$
Expanding absolute value:
$\displaystyle x > 1 \wedge {\color{Red} \sqrt{x-1} \geq x} \vee x \leq 1 \wedge \sqrt{1-x} \geq x$
Squaring:
$\displaystyle x > 1 \wedge {\color{Red} x-1 \geq x^2} \vee x \leq 1 \wedge 1-x \geq x^2$
Sorting:
$\displaystyle x > 1 \wedge {\color{Red} x^2-x+1 \leq 0} \vee x \leq 1 \wedge x^2+x-1 \leq 0$
$\displaystyle x > 1 \wedge \text{{\color{Red} no x}} \vee x \leq 1 \wedge \frac{-\sqrt{5}-1}{2} \leq x \leq \frac{\sqrt{5}-1}{2}$
$\displaystyle x \leq \frac{\sqrt{5}-1}{2}$
$\displaystyle \sqrt{|1-x|} \geq x$ if $\displaystyle \left \{ x \in \mathbb{R}\mid x \leq \frac{\sqrt{5}-1}{2} \right \}$