Proving $|\sqrt{x}-1|\leqslant|x-1|$

Question

I tried squaring both sides, didn’t get me anywhere. Maybe going case by case would result in something but I think there could be amore elegant proof.

Answer 1

Hint: For $x \ge 0$,

$$x-1 = (\sqrt{x}-1)(\sqrt{x}+1)$$

Try to take absolute value on both sides.

Edit:

$$|x-1| = |\sqrt{x}-1||\sqrt{x}+1| \ge |\sqrt{x}-1|(1)=|\sqrt{x}-1|$$

Answer 2

Note that we have for $x,y>0$ $$|\sqrt{x}-\sqrt{y}| = \frac{|(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})|}{\sqrt{x}+\sqrt{y}} = \frac{|x-y|}{\sqrt{x}+\sqrt{y}}.$$ Thus, taking $y=1$, we find $$|\sqrt{x}-1| = \frac{|x-1|}{\sqrt{x}+1} \leq |x-1|.$$

Answer 3

Let $x\ge 0$:

$|x-1|= |√x-1||√x+1| \ge$

$ |√x-1|\cdot 1 =|√x-1|.$

Used: $|√x+1| \ge 1.$