Question-125960

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

Answered by mahdipoor last updated on 16/Dec/20

{\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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