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: $\displaystyle x > 1 \wedge {\color{Red} x^2-x+1 \leq 0} \vee x \leq 1 \wedge x^2+x-1 \leq 0$

Solving quad. ineq.: $\displaystyle x > 1 \wedge \text{{\color{Red} no x}} \vee x \leq 1 \wedge \frac{-\sqrt{5}-1}{2} \leq x \leq \frac{\sqrt{5}-1}{2}$

Combining: $\displaystyle x \leq \frac{\sqrt{5}-1}{2}$

Note: the inequality marked in red is never true. I thinks this OK because the given is also never true for x > 1.

Note 2: i don't know any symbol/notation for "no x".

Conclusion:

$\displaystyle \sqrt{|1-x|} \geq x$ if $\displaystyle \left \{ x \in \mathbb{R}\mid x \leq \frac{\sqrt{5}-1}{2} \right \}$

Does this make sense at all?