Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 136739 by Ar Brandon last updated on 25/Mar/21

Given 0<a<b, prove that (((b−a)^2 )/(8b))≤((a+b)/2)−(√(ab))≤(((b−a)^2 )/(8a))

Given0<a<b,provethat (b−a)28b⩽a+b2−ab⩽(b−a)28a

Answered by snipers237 last updated on 26/Mar/21

((a+b)/2)−(√(ab ))= ((((√a)−(√b))^2 )/2) and b−a=((√b)−(√a))((√b)+(√a)) 0<a<b⇒(√a)<(√b) . Then 2(√a) < (√a)+(√b) <2(√b) So 4a<((√a)+(√b))^2 <4b (((b−a)^2 )/(8b))=((((√b)−(√a))^2 )/2).((((√b)+(√a))^2 )/(4b))<((((√a)−(√b))^2 )/2) (((b−a)^2 )/(8a))=((((√b)−(√a))^2 )/2).((((√b)+(√a))^2 )/(4a)) >((((√a)−(√b))^2 )/2) LFYTC

a+b2−ab=(a−b)22andb−a=(b−a)(b+a) 0<a<b⇒a<b.Then2a<a+b<2b So4a<(a+b)2<4b (b−a)28b=(b−a)22.(b+a)24b<(a−b)22 (b−a)28a=(b−a)22.(b+a)24a>(a−b)22 LFYTC

Commented byAr Brandon last updated on 26/Mar/21

Thanks

Thanks

Terms of Service

Privacy Policy

Contact: info@tinkutara.com