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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.

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

3 Answers 3

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Hint: For $x \ge 0$,

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

Try to take absolute value on both sides.

Edit:

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

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  • $\begingroup$ taking absolute values here doesn’t change anything right? $\endgroup$
    – user500668
    Mar 9, 2018 at 9:38
  • $\begingroup$ What can you say about the lower bound of $|\sqrt{x}+1|=\sqrt{x}+1$? $\endgroup$ Mar 9, 2018 at 9:39
  • $\begingroup$ I think it’s zero? $\endgroup$
    – user500668
    Mar 9, 2018 at 9:57
  • $\begingroup$ $0$ is indeed a lower bound but actually we know a little bit more. $\sqrt{x} \ge 0$, what about $\sqrt{x}+1$? $\endgroup$ Mar 9, 2018 at 10:06
  • $\begingroup$ lower bound of $\sqrt{x}+1$ is 1, so? $\endgroup$
    – user500668
    Mar 9, 2018 at 10:24
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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 $$|\sqrt{x}-1| = \frac{|x-1|}{\sqrt{x}+1} \leq |x-1|.$$

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Let $x\ge 0$:

$|x-1|= |√x-1||√x+1| \ge$

$ |√x-1|\cdot 1 =|√x-1|.$

Used: $|√x+1| \ge 1.$

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