prove-that-x-y-x-y-

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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