Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

show that ∣x+y∣≤∣x∣+∣y∣

showthat∣x+y∣⩽∣x∣+∣y∣

Answered by Bird last updated on 21/Oct/20

(∣x∣+∣y∣)^2 −∣x+y∣^2 = x^2 +2∣xy∣+y^2 −x^2 −2xy−y^2 =2(∣xy∣−xy)≥0 due to xy≤∣xy∣ ∣x∣+∣y∣≥0 and ∣x+y∣≥0 ⇒ ∣x+y∣≤∣x∣+∣y∣

(∣x∣+∣y∣)2−∣x+y∣2=x2+2∣xy∣+y2−x2−2xy−y2=2(∣xy∣−xy)⩾0duetoxy⩽∣xy∣∣x∣+∣y∣⩾0and∣x+y∣⩾0⇒∣x+y∣⩽∣x∣+∣y∣

Commented by mathocean1 last updated on 25/Oct/20

Thank you all sirs.

Thankyouallsirs.

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∣⇒a+b≤∣a∣+b≤∣a∣+∣b∣ a≥−∣a∣⇒a+b≥b−∣a∣≥−∣b∣−∣a∣ ⇒a+b≥−(∣a∣+∣b∣) ⇒−(∣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.

step1.lemma:−∣x∣⩽x⩽∣x∣proof:case1.ifx⩾0thenx=∣x∣⩽∣x∣butalsox⩾0⩾−∣x∣.case2.ifx⩽0thenx⩽0⩽∣x∣butx=−(−x)=−∣x∣⩾−∣x∣step2.bylemma:−∣a∣⩽a⩽∣a∣a⩽∣a∣⇒a+b⩽∣a∣+b⩽∣a∣+∣b∣a⩾−∣a∣⇒a+b⩾b−∣a∣⩾−∣b∣−∣a∣⇒a+b⩾−(∣a∣+∣b∣)⇒−(∣a∣+∣b∣)⩽a+b⩽∣a∣+∣b∣step3.toshow∣a+b∣⩽∣a∣+∣b∣dobycases:case1.a+b⩾0then∣a+b∣=a+b⩽∣a∣+∣b∣.case2a+b⩽0then∣a+b∣=−(a+b)⩽−(−(∣a∣+∣b∣))=∣a∣+∣b∣.Proved.

Commented by mathocean1 last updated on 25/Oct/20

Thanks.

Thanks.

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

∣x+y∣≤∣x∣+∣y∣⇔∣x+y∣^2 ≤(∣x∣+∣y∣)^2 ⇔x^2 +2xy+y^2 ≤x^2 +2∣xy∣+y^2 ⇔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)

∣x+y∣⩽∣x∣+∣y∣⇔∣x+y∣2⩽(∣x∣+∣y∣)2⇔x2+2xy+y2⩽x2+2∣xy∣+y2⇔xy⩽∣xy∣.Thelastinequalityisalwaystrue,sothegiveninequalityistrue.Theequalityocurrsifandonlyifxy⩾0(q.e.d)

Commented by mathocean1 last updated on 25/Oct/20

thanks.

thanks.

Terms of Service

Privacy Policy

Contact: info@tinkutara.com