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Question Number 1543 by 112358 last updated on 17/Aug/15

Let a, b∈R. Show that                  ∣∣a∣−∣b∣∣≤∣a−b∣.

$${Let}\:{a},\:{b}\in\mathbb{R}.\:{Show}\:{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mid{a}\mid−\mid{b}\mid\mid\leqslant\mid{a}−{b}\mid. \\ $$

Answered by Rasheed Soomro last updated on 18/Aug/15

Alternative way , technically simpler.  ∣∣a∣−∣b∣∣≤∣a−b∣......................(I)  Let ∣a∣ = x    and      ∣b∣ = y   ,  Clearly  x,y≥0  ∣a∣=x ⇒ a=±x   and     ∣b∣=±y  By substituting in (I) we have four simpler  inequalities:           ∣ x−y ∣≤ ∣ (+x)−(+y) ∣= ∣ x−y ∣......................(II)                         ≤ ∣ (+x)−(−y) ∣=∣ x+y ∣.......(III)                          ≤∣ (−x)−(+y) ∣=∣−(x+y)∣= ∣ x+y ∣..........(IV)                          ≤∣ (−x)−(−y) ∣=∣ −(x−y) ∣= ∣ x−y ∣ .....(V)  (II)   and  (V)    are obvious.For (III) and (IV) we  proceed as follow           x−y≤x+y   for x and y are both +ve.     ∴  ∣ x−y ∣ ≤ ∣x + y∣  Since (II) ,  (III) , (IV) and (V) are true. It implies that                      ∣∣a∣−∣b∣∣≤∣a−b∣

$$\boldsymbol{\mathrm{Alternative}}\:\boldsymbol{\mathrm{way}}\:,\:\boldsymbol{\mathrm{technically}}\:\boldsymbol{\mathrm{simpler}}. \\ $$$$\mid\mid{a}\mid−\mid{b}\mid\mid\leqslant\mid{a}−{b}\mid......................\left(\boldsymbol{\mathrm{I}}\right) \\ $$$$\mathrm{L}{et}\:\mid{a}\mid\:=\:{x}\:\:\:\:{and}\:\:\:\:\:\:\mid{b}\mid\:=\:{y}\:\:\:,\:\:{Clearly}\:\:{x},{y}\geqslant\mathrm{0} \\ $$$$\mid{a}\mid={x}\:\Rightarrow\:{a}=\pm{x}\:\:\:{and}\:\:\:\:\:\mid{b}\mid=\pm{y} \\ $$$${By}\:{substituting}\:{in}\:\left(\boldsymbol{\mathrm{I}}\right)\:{we}\:{have}\:{four}\:{simpler}\:\:{inequalities}: \\ $$$$\:\:\:\:\:\:\:\:\:\mid\:{x}−{y}\:\mid\leqslant\:\mid\:\left(+{x}\right)−\left(+{y}\right)\:\mid=\:\mid\:{x}−{y}\:\mid......................\left(\boldsymbol{\mathrm{II}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leqslant\:\mid\:\left(+{x}\right)−\left(−{y}\right)\:\mid=\mid\:{x}+{y}\:\mid.......\left(\boldsymbol{\mathrm{III}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leqslant\mid\:\left(−{x}\right)−\left(+{y}\right)\:\mid=\mid−\left({x}+{y}\right)\mid=\:\mid\:{x}+{y}\:\mid..........\left(\boldsymbol{\mathrm{IV}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leqslant\mid\:\left(−{x}\right)−\left(−{y}\right)\:\mid=\mid\:−\left({x}−{y}\right)\:\mid=\:\mid\:{x}−{y}\:\mid\:.....\left(\boldsymbol{\mathrm{V}}\right) \\ $$$$\left(\boldsymbol{\mathrm{II}}\right)\:\:\:{and}\:\:\left(\boldsymbol{\mathrm{V}}\right)\:\:\:\:{are}\:{obvious}.{For}\:\left(\boldsymbol{\mathrm{III}}\right)\:{and}\:\left(\boldsymbol{\mathrm{IV}}\right)\:{we} \\ $$$${proceed}\:{as}\:{follow} \\ $$$$\:\:\:\:\:\:\:\:\:{x}−{y}\leqslant{x}+{y}\:\:\:{for}\:{x}\:{and}\:{y}\:{are}\:{both}\:+{ve}. \\ $$$$\:\:\:\therefore\:\:\mid\:{x}−{y}\:\mid\:\leqslant\:\mid{x}\:+\:{y}\mid \\ $$$${Since}\:\left(\boldsymbol{\mathrm{II}}\right)\:,\:\:\left(\boldsymbol{\mathrm{III}}\right)\:,\:\left(\boldsymbol{\mathrm{IV}}\right)\:{and}\:\left(\boldsymbol{\mathrm{V}}\right)\:{are}\:{true}.\:{It}\:{implies}\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mid{a}\mid−\mid{b}\mid\mid\leqslant\mid{a}−{b}\mid \\ $$

Answered by 123456 last updated on 17/Aug/15

we have that  0≤∣a+b∣≤∣a∣+∣b∣  0≤∣a+(b−a)∣≤∣a∣+∣b−a∣⇒−∣a∣≤∣b−a∣  0≤∣b∣≤∣a∣+∣b−a∣⇒∣a∣≤∣a+b∣≤∣a∣+∣b∣≤2∣a∣+∣b−a∣  −∣a∣≤∣b∣−∣a∣≤∣b−a∣  we have  −∣a∣≤∣a∣≤∣b−a∣⇒−∣b−a∣≤−∣a∣≤∣a∣  −∣b−a∣≤−∣a∣≤∣b∣−∣a∣≤∣b−a∣  ∣∣b∣−∣a∣∣≤∣b−a∣

$$\mathrm{we}\:\mathrm{have}\:\mathrm{that} \\ $$$$\mathrm{0}\leqslant\mid{a}+{b}\mid\leqslant\mid{a}\mid+\mid{b}\mid \\ $$$$\mathrm{0}\leqslant\mid{a}+\left({b}−{a}\right)\mid\leqslant\mid{a}\mid+\mid{b}−{a}\mid\Rightarrow−\mid{a}\mid\leqslant\mid{b}−{a}\mid \\ $$$$\mathrm{0}\leqslant\mid{b}\mid\leqslant\mid{a}\mid+\mid{b}−{a}\mid\Rightarrow\mid{a}\mid\leqslant\mid{a}+{b}\mid\leqslant\mid{a}\mid+\mid{b}\mid\leqslant\mathrm{2}\mid{a}\mid+\mid{b}−{a}\mid \\ $$$$−\mid{a}\mid\leqslant\mid{b}\mid−\mid{a}\mid\leqslant\mid{b}−{a}\mid \\ $$$$\mathrm{we}\:\mathrm{have} \\ $$$$−\mid{a}\mid\leqslant\mid{a}\mid\leqslant\mid{b}−{a}\mid\Rightarrow−\mid{b}−{a}\mid\leqslant−\mid{a}\mid\leqslant\mid{a}\mid \\ $$$$−\mid{b}−{a}\mid\leqslant−\mid{a}\mid\leqslant\mid{b}\mid−\mid{a}\mid\leqslant\mid{b}−{a}\mid \\ $$$$\mid\mid{b}\mid−\mid{a}\mid\mid\leqslant\mid{b}−{a}\mid \\ $$

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