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Question Number 125960 by deleteduser1 last updated on 26/Sep/22


	Answered by mahdipoor last updated on 16/Dec/20
	${ ((a^2 +b^2 =∣a∣^2 +∣b∣^2 )),((2ab≤2∣a∣.∣b∣)) :} ⇒a^2 +b^2 +2ab≤∣a∣^2 +∣b∣^2 +2∣a∣.∣b∣ ⇒(a+b)^2 ≤(∣a∣+∣b∣)^2 ⇒∣a+b∣≤∣a∣+∣b∣$
	${\begin{cases} a^{2} + b^{2} =∣ a ∣^{2} + ∣ b ∣^{2} \\ 2 a b ⩽ 2 ∣ a ∣ . ∣ b ∣ \end{cases}$ $\Rightarrow a^{2} + b^{2} + 2 a b ⩽∣ a ∣^{2} + ∣ b ∣^{2} + 2 ∣ a ∣ . ∣ b ∣$ $\Rightarrow {(a + b)}^{2} ⩽ {(∣ a ∣ + ∣ b ∣)}^{2}$ $\Rightarrow∣ a + b ∣⩽∣ a ∣ + ∣ b ∣$
	Commented by mahdipoor last updated on 16/Dec/20

	$x^{2} ⩽ y^{2} \Rightarrow∣ x ∣⩽∣ y ∣$
	Commented by mahdipoor last updated on 16/Dec/20

	$i n m y a n s; x = a + b y =∣ a ∣ + ∣ b ∣$ $∣ x ∣=∣ a + b ∣ ∣ y ∣=∣∣ a ∣ + ∣ b ∣∣=∣ a ∣ + ∣ b ∣$
	Answered by arcana last updated on 19/Dec/20

	$using - ∣ a ∣⩽ a ⩽∣ a ∣ and$ $- ∣ b ∣⩽ b ⩽∣ b ∣,$ $then$ $- ∣ a ∣ - ∣ b ∣⩽ a + b ⩽∣ a ∣ + ∣ b ∣$ $- (∣ a ∣ + ∣ b ∣) ⩽ a + b ⩽∣ a ∣ + ∣ b ∣$ $\Rightarrow$ $∣ a + b ∣⩽∣ a ∣ + ∣ b ∣$
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