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Question Number 164622 by mathocean1 last updated on 19/Jan/22

Show that ∀ a, b ∈ R, 1. ∣∣x∣−∣y∣∣≤∣x−y∣ 2. 1+∣xy−1∣≤(1+∣x−1∣)(1+∣y−1∣).

$${Show}\:{that}\:\forall\:{a},\:{b}\:\in\:\mathbb{R}, \\ $$$$\mathrm{1}.\:\mid\mid{x}\mid−\mid{y}\mid\mid\leqslant\mid{x}−{y}\mid \\ $$$$\mathrm{2}.\:\mathrm{1}+\mid{xy}−\mathrm{1}\mid\leqslant\left(\mathrm{1}+\mid{x}−\mathrm{1}\mid\right)\left(\mathrm{1}+\mid{y}−\mathrm{1}\mid\right). \\ $$

Answered by hmrsh last updated on 19/Jan/22

Answer of part(1): ∗ ∣x∣ ≤ A ↔ −A ≤ x ≤ A ∗∗ ∣x + y∣ ≤ ∣x∣ + ∣y∣ (triangle inequality) ∣x∣ = ∣(x − y) + y∣ →^(∗∗) ∣x∣ = ∣(x − y) + y∣ ≤ ∣x − y∣ + ∣y∣ (#) also: ∣y∣ = ∣(y−x)+x∣ ≤ ∣y−x∣+∣x∣=∣x−y∣+∣x∣ → ∣y∣ −∣x − y∣ ≤ ∣x∣ →^((#)) ∣y∣ −∣x − y∣ ≤ ∣x∣ ≤ ∣x − y∣ + ∣y∣ → −∣x − y∣ ≤ ∣x∣ − ∣y∣ ≤ ∣x − y∣ →^∗ ∣∣x∣ − ∣y∣∣ ≤ ∣x − y∣

$$ \\ $$$$\mathrm{Answer}\:\mathrm{of}\:\mathrm{part}\left(\mathrm{1}\right): \\ $$$$\ast\:\mid{x}\mid\:\leqslant\:{A}\:\leftrightarrow\:−{A}\:\leqslant\:{x}\:\leqslant\:{A} \\ $$$$\ast\ast\:\mid{x}\:+\:{y}\mid\:\leqslant\:\mid{x}\mid\:+\:\mid{y}\mid\:\:\:\:\:\:\left(\mathrm{triangle}\:\mathrm{inequality}\right) \\ $$$$ \\ $$$$\mid{x}\mid\:=\:\mid\left({x}\:−\:{y}\right)\:+\:{y}\mid \\ $$$$\overset{\ast\ast} {\rightarrow}\:\mid{x}\mid\:=\:\mid\left({x}\:−\:{y}\right)\:+\:{y}\mid\:\leqslant\:\mid{x}\:−\:{y}\mid\:+\:\mid{y}\mid\:\:\:\left(#\right) \\ $$$$\mathrm{also}: \\ $$$$\mid{y}\mid\:=\:\mid\left({y}−{x}\right)+{x}\mid\:\leqslant\:\mid{y}−{x}\mid+\mid{x}\mid=\mid{x}−{y}\mid+\mid{x}\mid \\ $$$$\rightarrow\:\mid{y}\mid\:−\mid{x}\:−\:{y}\mid\:\leqslant\:\mid{x}\mid \\ $$$$ \\ $$$$\overset{\left(#\right)} {\rightarrow}\:\mid{y}\mid\:−\mid{x}\:−\:{y}\mid\:\leqslant\:\mid{x}\mid\:\leqslant\:\mid{x}\:−\:{y}\mid\:+\:\mid{y}\mid \\ $$$$\rightarrow\:−\mid{x}\:−\:{y}\mid\:\leqslant\:\mid{x}\mid\:−\:\mid{y}\mid\:\leqslant\:\mid{x}\:−\:{y}\mid \\ $$$$\overset{\ast} {\rightarrow}\:\mid\mid{x}\mid\:−\:\mid{y}\mid\mid\:\leqslant\:\mid{x}\:−\:{y}\mid \\ $$

Commented by mathocean1 last updated on 19/Jan/22

thanks.

$${thanks}. \\ $$

Answered by hmrsh last updated on 20/Jan/22

Answer of part 2: tip 1: xy−1 = (1−x)(1−y)+(x−1)+(y−1) tip 2: (1+x_1 )(1+x_2 ) = 1+ x_1 +x_2 +x_1 x_2 using triangle inequality and tip 1: ∣xy−1∣ = ∣(1−x)(1−y)+(x−1)+(y−1)∣ ≤ ∣(1−x)(1−y)∣+∣(x−1)+(y−1)∣ ≤ ∣(1−x)(1−y)∣+∣x−1∣+∣y−1∣ = ∣1−x∣∣1−y∣+∣x−1∣+∣y−1∣ = ∣x−1∣∣y−1∣+∣x−1∣+∣y−1∣ → ∣xy−1∣≤∣x−1∣∣y−1∣+∣x−1∣+∣y−1∣ →^(+1) ∣xy−1∣+1 ≤∣x−1∣∣y−1∣+∣x−1∣+∣y−1∣+1 using tip 2 to RHS: ∣xy−1∣+1≤(1+∣x−1∣)(1+∣y−1∣)

$$\mathrm{Answer}\:\mathrm{of}\:\mathrm{part}\:\mathrm{2}: \\ $$$$\mathrm{tip}\:\mathrm{1}: \\ $$$${xy}−\mathrm{1}\:=\:\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)+\left({x}−\mathrm{1}\right)+\left({y}−\mathrm{1}\right) \\ $$$$\mathrm{tip}\:\mathrm{2}: \\ $$$$\left(\mathrm{1}+{x}_{\mathrm{1}} \right)\left(\mathrm{1}+{x}_{\mathrm{2}} \right)\:=\:\mathrm{1}+\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{1}} {x}_{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{using}\:\mathrm{triangle}\:\mathrm{inequality}\:\mathrm{and}\:\mathrm{tip}\:\mathrm{1}: \\ $$$$\mid{xy}−\mathrm{1}\mid\:=\:\mid\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)+\left({x}−\mathrm{1}\right)+\left({y}−\mathrm{1}\right)\mid \\ $$$$\leqslant\:\mid\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)\mid+\mid\left({x}−\mathrm{1}\right)+\left({y}−\mathrm{1}\right)\mid \\ $$$$\leqslant\:\mid\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)\mid+\mid{x}−\mathrm{1}\mid+\mid{y}−\mathrm{1}\mid \\ $$$$=\:\mid\mathrm{1}−{x}\mid\mid\mathrm{1}−{y}\mid+\mid{x}−\mathrm{1}\mid+\mid{y}−\mathrm{1}\mid \\ $$$$=\:\mid{x}−\mathrm{1}\mid\mid{y}−\mathrm{1}\mid+\mid{x}−\mathrm{1}\mid+\mid{y}−\mathrm{1}\mid \\ $$$$ \\ $$$$\rightarrow\:\mid{xy}−\mathrm{1}\mid\leqslant\mid{x}−\mathrm{1}\mid\mid{y}−\mathrm{1}\mid+\mid{x}−\mathrm{1}\mid+\mid{y}−\mathrm{1}\mid \\ $$$$\overset{+\mathrm{1}} {\rightarrow}\:\mid{xy}−\mathrm{1}\mid+\mathrm{1}\:\leqslant\mid{x}−\mathrm{1}\mid\mid{y}−\mathrm{1}\mid+\mid{x}−\mathrm{1}\mid+\mid{y}−\mathrm{1}\mid+\mathrm{1} \\ $$$$\mathrm{using}\:\mathrm{tip}\:\mathrm{2}\:\mathrm{to}\:\mathrm{RHS}: \\ $$$$\mid{xy}−\mathrm{1}\mid+\mathrm{1}\leqslant\left(\mathrm{1}+\mid{x}−\mathrm{1}\mid\right)\left(\mathrm{1}+\mid{y}−\mathrm{1}\mid\right) \\ $$$$ \\ $$

Commented by mathocean1 last updated on 22/Jan/22

thanks sir.

$${thanks}\:{sir}. \\ $$

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