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

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.

• Do you need to prove the inequality for all $x \ge 0$? Mar 9, 2018 at 9:23

Hint: For $x \ge 0$,

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

Try to take absolute value on both sides.

Edit:

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

• taking absolute values here doesn’t change anything right? Mar 9, 2018 at 9:38
• What can you say about the lower bound of $|\sqrt{x}+1|=\sqrt{x}+1$? Mar 9, 2018 at 9:39
• I think it’s zero? Mar 9, 2018 at 9:57
• $0$ is indeed a lower bound but actually we know a little bit more. $\sqrt{x} \ge 0$, what about $\sqrt{x}+1$? Mar 9, 2018 at 10:06
• lower bound of $\sqrt{x}+1$ is 1, so? Mar 9, 2018 at 10:24

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 $$|\sqrt{x}-1| = \frac{|x-1|}{\sqrt{x}+1} \leq |x-1|.$$

Let $x\ge 0$:

$|x-1|= |√x-1||√x+1| \ge$

$|√x-1|\cdot 1 =|√x-1|.$

Used: $|√x+1| \ge 1.$