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Question Number 71082 by aliesam last updated on 11/Oct/19
prove that ∣ (√(∣x∣)) − (√(∣y∣)) ∣ ≤ (√(∣x−y∣))
$${prove}\:{that}\: \\ $$$$ \\ $$$$\mid\:\sqrt{\mid{x}\mid}\:−\:\sqrt{\mid{y}\mid}\:\mid\:\leqslant\:\sqrt{\mid{x}−{y}\mid}\: \\ $$$$ \\ $$
Answered by Henri Boucatchou last updated on 11/Oct/19
As ∣x∣ − ∣y∣ ≤ ∣∣x∣ − ∣y∣∣ ≤ ∣x − y∣, (√(∣x∣)) − (√(∣y∣)) ≤ ∣(√(∣x∣)) − (√(∣y∣))∣ If ∣(√(∣x∣)) − (√(∣y∣))∣ > (√(∣x − y∣)) , take x=4 and y=9 ⇒ ∣(√4) − (√9)∣ = ∣2−3∣ = 1 > (√(∣4−9∣)) = (√5) : absurd; so ∣(√(∣x∣)) − (√(∣y∣))∣ ≤ (√(∣x − y∣)).
$${As}\:\:\mid{x}\mid\:−\:\mid{y}\mid\:\leqslant\:\mid\mid{x}\mid\:−\:\mid{y}\mid\mid\:\leqslant\:\mid{x}\:−\:{y}\mid, \\ $$$$\:\:\sqrt{\mid{x}\mid}\:−\:\sqrt{\mid{y}\mid}\:\leqslant\:\mid\sqrt{\mid{x}\mid}\:−\:\sqrt{\mid{y}\mid}\mid\:\: \\ $$$$\:\:{If}\:\:\mid\sqrt{\mid{x}\mid}\:−\:\sqrt{\mid{y}\mid}\mid\:>\:\sqrt{\mid{x}\:−\:{y}\mid}\:,\:\:{take}\:\:{x}=\mathrm{4}\:\:{and}\:\:{y}=\mathrm{9}\:\Rightarrow\:\:\mid\sqrt{\mathrm{4}}\:−\:\sqrt{\mathrm{9}}\mid\:=\:\mid\mathrm{2}−\mathrm{3}\mid\:=\:\mathrm{1}\:>\:\sqrt{\mid\mathrm{4}−\mathrm{9}\mid}\:=\:\sqrt{\mathrm{5}}\::\:\:{absurd}; \\ $$$$\:\:{so}\:\:\mid\sqrt{\mid{x}\mid}\:−\:\sqrt{\mid{y}\mid}\mid\:\leqslant\:\sqrt{\mid{x}\:−\:{y}\mid}. \\ $$
Commented by aliesam last updated on 11/Oct/19
can you prove that without using numbers and thank you for that sol
$${can}\:{you}\:{prove}\:{that}\:{without}\:{using}\:{numbers}\:{and}\:{thank}\:{you}\:{for}\:{that}\:{sol} \\ $$
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 ⇒we can so reduce possibilty st x≥y≥0 ⇒(√x)≥(√y) ⇒−2(√(xy))≤−2y ⇒x+y−2(√(xy))≤x+y−2y ⇒((√x)−(√y))^2 ≤(x−y)=((√(x−y)))^2 ⇒(√x)−(√y)≤(√(x−y)) ∴ since xy≥0 we are (x≥0,y≥0)∪(x≤0,y≤0) change (x,y) and (−x,−y)⇒R(x,y)elation is symetric over ofigine so we can reduce (x≥0,y≥0) change (x,y) and (y,x) ⇒relation symetric over y=x⇒R(x,y)∣x≥y⇔R(x,y)∣y≤x
$$\mathrm{if}\:\mathrm{xy}\leqslant\mathrm{0}\:\:\mathrm{its}\:\:\:\mathrm{clear} \\ $$$$\mathrm{xy}\geqslant\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{switch}\:\:\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{by}\:\left(\mathrm{y},\mathrm{x}\right)\:\mathrm{without}\:\mathrm{changing}\:\mathrm{inqyality} \\ $$$$\mathrm{and}\:\mathrm{change}\:\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{withe}\:\left(−\mathrm{x},−\mathrm{y}\right)\:\mathrm{withoute}\:\mathrm{changing}\:\mathrm{the}\:\mathrm{inquality} \\ $$$$\Rightarrow\mathrm{we}\:\mathrm{can}\:\:\mathrm{so}\:\mathrm{reduce}\:\mathrm{possibilty}\:\mathrm{st} \\ $$$$\mathrm{x}\geqslant\mathrm{y}\geqslant\mathrm{0} \\ $$$$\Rightarrow\sqrt{\mathrm{x}}\geqslant\sqrt{\mathrm{y}} \\ $$$$\Rightarrow−\mathrm{2}\sqrt{\mathrm{xy}}\leqslant−\mathrm{2y} \\ $$$$\Rightarrow\mathrm{x}+\mathrm{y}−\mathrm{2}\sqrt{\mathrm{xy}}\leqslant\mathrm{x}+\mathrm{y}−\mathrm{2y} \\ $$$$\Rightarrow\left(\sqrt{\mathrm{x}}−\sqrt{\mathrm{y}}\right)^{\mathrm{2}} \leqslant\left(\mathrm{x}−\mathrm{y}\right)=\left(\sqrt{\mathrm{x}−\mathrm{y}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\mathrm{x}}−\sqrt{\mathrm{y}}\leqslant\sqrt{\mathrm{x}−\mathrm{y}} \\ $$$$\therefore\:\mathrm{since}\:\mathrm{xy}\geqslant\mathrm{0}\:\mathrm{we}\:\mathrm{are}\:\:\left(\mathrm{x}\geqslant\mathrm{0},\mathrm{y}\geqslant\mathrm{0}\right)\cup\left(\mathrm{x}\leqslant\mathrm{0},\mathrm{y}\leqslant\mathrm{0}\right) \\ $$$$\mathrm{change}\:\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{and}\:\left(−\mathrm{x},−\mathrm{y}\right)\Rightarrow\mathrm{R}\left(\mathrm{x},\mathrm{y}\right)\mathrm{elation}\:\mathrm{is}\:\mathrm{symetric}\:\mathrm{over}\:\mathrm{ofigine} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{can}\:\mathrm{reduce}\:\left(\mathrm{x}\geqslant\mathrm{0},\mathrm{y}\geqslant\mathrm{0}\right)\:\mathrm{change}\:\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{and}\:\left(\mathrm{y},\mathrm{x}\right)\:\Rightarrow\mathrm{relation} \\ $$$$\mathrm{symetric}\:\mathrm{over}\:\mathrm{y}=\mathrm{x}\Rightarrow\mathrm{R}\left(\mathrm{x},\mathrm{y}\right)\mid\mathrm{x}\geqslant\mathrm{y}\Leftrightarrow\mathrm{R}\left(\mathrm{x},\mathrm{y}\right)\mid\mathrm{y}\leqslant\mathrm{x} \\ $$$$ \\ $$
Commented by aliesam last updated on 11/Oct/19
thank you sir great work
$${thank}\:{you}\:{sir}\:{great}\:{work} \\ $$

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