show-that-x-y-x-y-

Question Number 119057 by mathocean1 last updated on 21/Oct/20

s h o w t h a t

∣ x + y ∣⩽∣ x ∣ + ∣ y ∣

Answered by Bird last updated on 21/Oct/20

{(∣ x ∣ + ∣ y ∣)}^{2} - ∣ x + y ∣^{2} =

x^{2} + 2 ∣ x y ∣ + y^{2} - x^{2} - 2 x y - y^{2}

= 2 (∣ x y ∣ - x y) ⩾ 0 d u e t o x y ⩽∣ x y ∣

∣ x ∣ + ∣ y ∣⩾ 0 a n d ∣ x + y ∣⩾ 0 \Rightarrow

∣ x + y ∣⩽∣ x ∣ + ∣ y ∣

Commented by mathocean1 last updated on 25/Oct/20

Thank you all sirs .

Answered by floor(10²Eta[1]) last updated on 22/Oct/20

step 1 .

lemma :

- ∣ x ∣⩽ x ⩽∣ x ∣

proof :

case 1 .

if x ⩾ 0 then x =∣ x ∣⩽∣ x ∣

but also x ⩾ 0 ⩾ - ∣ x ∣ .

case 2 .

if x ⩽ 0 then x ⩽ 0 ⩽∣ x ∣

but x = - (- x) = - ∣ x ∣⩾ - ∣ x ∣

step 2 .

by lemma : - ∣ a ∣⩽ a ⩽∣ a ∣

a ⩽∣ a ∣\Rightarrow a + b ⩽∣ a ∣ + b ⩽∣ a ∣ + ∣ b ∣

a ⩾ - ∣ a ∣\Rightarrow a + b ⩾ b - ∣ a ∣⩾ - ∣ b ∣ - ∣ a ∣

\Rightarrow a + b ⩾ - (∣ a ∣ + ∣ b ∣)

\Rightarrow - (∣ a ∣ + ∣ b ∣) ⩽ a + b ⩽∣ a ∣ + ∣ b ∣

step 3 .

to show ∣ a + b ∣⩽∣ a ∣ + ∣ b ∣ do by cases :

case 1 .

a + b ⩾ 0 then ∣ a + b ∣= a + b ⩽∣ a ∣ + ∣ b ∣ .

case 2

a + b ⩽ 0 then

∣ a + b ∣= - (a + b) ⩽ - (- (∣ a ∣ + ∣ b ∣)) =∣ a ∣ + ∣ b ∣ .

Proved .

Commented by mathocean1 last updated on 25/Oct/20

Thanks .

Answered by 1549442205PVT last updated on 22/Oct/20

∣ x + y ∣⩽∣ x ∣ + ∣ y ∣\Leftrightarrow∣ x + y ∣^{2} ⩽ {(∣ x ∣ + ∣ y ∣)}^{2}

\Leftrightarrow x^{2} + 2 xy + y^{2} ⩽ x^{2} + 2 ∣ xy ∣ + y^{2}

\Leftrightarrow xy ⩽∣ xy ∣ . The last inequality is always

true, so the given inequality is true

. The equality ocurrs if and only if

xy ⩾ 0 (q . e . d)

Commented by mathocean1 last updated on 25/Oct/20

thanks .