prove that ∣ (√(∣x∣)) − (√(∣y∣)) ∣ ≤ (√(∣x−y∣))


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Question Number 71082 by aliesam last updated on 11/Oct/19

$p r o v e t h a t$ $∣ \sqrt{∣ x ∣} - \sqrt{∣ y ∣} ∣ ⩽ \sqrt{∣ x - y ∣}$
	Answered by Henri Boucatchou last updated on 11/Oct/19

	$A s ∣ x ∣ - ∣ y ∣ ⩽ ∣∣ x ∣ - ∣ y ∣∣ ⩽ ∣ x - y ∣,$ $\sqrt{∣ x ∣} - \sqrt{∣ y ∣} ⩽ ∣ \sqrt{∣ x ∣} - \sqrt{∣ y ∣} ∣$ $I f ∣ \sqrt{∣ x ∣} - \sqrt{∣ y ∣} ∣ > \sqrt{∣ x - y ∣}, t a k e x = 4 a n d y = 9 \Rightarrow ∣ \sqrt{4} - \sqrt{9} ∣ = ∣ 2 - 3 ∣ = 1 > \sqrt{∣ 4 - 9 ∣} = \sqrt{5} : a b s u r d;$ $s o ∣ \sqrt{∣ x ∣} - \sqrt{∣ y ∣} ∣ ⩽ \sqrt{∣ x - y ∣} .$
	Commented by aliesam last updated on 11/Oct/19

	$c a n y o u p r o v e t h a t w i t h o u t u s i n g n u m b e r s a n d t h a n k y o u f o r t h a t s o l$
	Answered by mind is power last updated on 11/Oct/19

	$if xy ⩽ 0 its clear$ $xy ⩾ 0$ $we can switch (x, y) by (y, x) without changing inqyality$ $and change (x, y) withe (- x, - y) withoute changing the inquality$ $\Rightarrow we can so reduce possibilty st$ $x ⩾ y ⩾ 0$ $\Rightarrow \sqrt{x} ⩾ \sqrt{y}$ $\Rightarrow - 2 \sqrt{xy} ⩽ - 2 y$ $\Rightarrow x + y - 2 \sqrt{xy} ⩽ x + y - 2 y$ $\Rightarrow {(\sqrt{x} - \sqrt{y})}^{2} ⩽ (x - y) = {(\sqrt{x - y})}^{2}$ $\Rightarrow \sqrt{x} - \sqrt{y} ⩽ \sqrt{x - y}$ $∴ since xy ⩾ 0 we are (x ⩾ 0, y ⩾ 0) \cup (x ⩽ 0, y ⩽ 0)$ $change (x, y) and (- x, - y) \Rightarrow R (x, y) elation is symetric over ofigine$ $so we can reduce (x ⩾ 0, y ⩾ 0) change (x, y) and (y, x) \Rightarrow relation$ $symetric over y = x \Rightarrow R (x, y) ∣ x ⩾ y \Leftrightarrow R (x, y) ∣ y ⩽ x$
	Commented by aliesam last updated on 11/Oct/19

	$t h a n k y o u s i r g r e a t w o r k$
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