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


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

$S h o w t h a t \forall a, b \in R,$ $1 . ∣∣ x ∣ - ∣ y ∣∣⩽∣ x - y ∣$ $2 . 1 + ∣ x y - 1 ∣⩽ (1 + ∣ x - 1 ∣) (1 + ∣ y - 1 ∣) .$
	Answered by hmrsh last updated on 19/Jan/22

	$Answer of part (1) :$ $* ∣ x ∣ ⩽ A \leftrightarrow - A ⩽ x ⩽ A$ $* * ∣ x + y ∣ ⩽ ∣ x ∣ + ∣ y ∣ (triangle inequality)$ $∣ x ∣ = ∣ (x - y) + y ∣$ $You can't use 'macro parameter character #' in math mode$ $also :$ $∣ y ∣ = ∣ (y - x) + x ∣ ⩽ ∣ y - x ∣ + ∣ x ∣=∣ x - y ∣ + ∣ x ∣$ $\to ∣ y ∣ - ∣ x - y ∣ ⩽ ∣ x ∣$ $You can't use 'macro parameter character #' in math mode$ $\to - ∣ x - y ∣ ⩽ ∣ x ∣ - ∣ y ∣ ⩽ ∣ x - y ∣$ $\overset{*}{\to} ∣∣ x ∣ - ∣ y ∣∣ ⩽ ∣ x - y ∣$
	Commented by mathocean1 last updated on 19/Jan/22

	$t h a n k s .$
	Answered by hmrsh last updated on 20/Jan/22

	$Answer of part 2 :$ $tip 1 :$ $x y - 1 = (1 - x) (1 - y) + (x - 1) + (y - 1)$ $tip 2 :$ $(1 + x_{1}) (1 + x_{2}) = 1 + x_{1} + x_{2} + x_{1} x_{2}$ $using triangle inequality and tip 1 :$ $∣ x y - 1 ∣ = ∣ (1 - x) (1 - y) + (x - 1) + (y - 1) ∣$ $⩽ ∣ (1 - x) (1 - y) ∣ + ∣ (x - 1) + (y - 1) ∣$ $⩽ ∣ (1 - x) (1 - y) ∣ + ∣ x - 1 ∣ + ∣ y - 1 ∣$ $= ∣ 1 - x ∣∣ 1 - y ∣ + ∣ x - 1 ∣ + ∣ y - 1 ∣$ $= ∣ x - 1 ∣∣ y - 1 ∣ + ∣ x - 1 ∣ + ∣ y - 1 ∣$ $\to ∣ x y - 1 ∣⩽∣ x - 1 ∣∣ y - 1 ∣ + ∣ x - 1 ∣ + ∣ y - 1 ∣$ $\overset{+ 1}{\to} ∣ x y - 1 ∣ + 1 ⩽∣ x - 1 ∣∣ y - 1 ∣ + ∣ x - 1 ∣ + ∣ y - 1 ∣ + 1$ $using tip 2 to RHS :$ $∣ x y - 1 ∣ + 1 ⩽ (1 + ∣ x - 1 ∣) (1 + ∣ y - 1 ∣)$
	Commented by mathocean1 last updated on 22/Jan/22

	$t h a n k s s i r .$
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