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 $x \ge 0$? $\endgroup$– Paolo LeonettiMar 9, 2018 at 9:23
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3 Answers
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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:
$$|x-1| = |\sqrt{x}-1||\sqrt{x}+1| \ge |\sqrt{x}-1|(1)=|\sqrt{x}-1|$$
answered Mar 9, 2018 at 9:27
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$\begingroup$ taking absolute values here doesn’t change anything right? $\endgroup$ Mar 9, 2018 at 9:38
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$\begingroup$ What can you say about the lower bound of $|\sqrt{x}+1|=\sqrt{x}+1$? $\endgroup$ Mar 9, 2018 at 9:39
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$\begingroup$ $0$ 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
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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: $|√x+1| \ge 1.$
answered Mar 9, 2018 at 10:19