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))

$$\mathrm{Given}\:\mathrm{0}<\mathrm{a}<\mathrm{b},\:\mathrm{prove}\:\mathrm{that} \\ $$ $$\frac{\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{2}} }{\mathrm{8b}}\leqslant\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}−\sqrt{\mathrm{ab}}\leqslant\frac{\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{2}} }{\mathrm{8a}} \\ $$

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

$$\frac{{a}+{b}}{\mathrm{2}}−\sqrt{{ab}\:}=\:\frac{\left(\sqrt{{a}}−\sqrt{{b}}\right)^{\mathrm{2}} }{\mathrm{2}}\:\:\:{and}\:\:{b}−{a}=\left(\sqrt{{b}}−\sqrt{{a}}\right)\left(\sqrt{{b}}+\sqrt{{a}}\right) \\ $$ $$\mathrm{0}<{a}<{b}\Rightarrow\sqrt{{a}}<\sqrt{{b}}\:.\:{Then}\:\:\mathrm{2}\sqrt{{a}}\:<\:\sqrt{{a}}+\sqrt{{b}}\:<\mathrm{2}\sqrt{{b}}\: \\ $$ $${So}\:\mathrm{4}{a}<\left(\sqrt{{a}}+\sqrt{{b}}\right)^{\mathrm{2}} <\mathrm{4}{b} \\ $$ $$\frac{\left({b}−{a}\right)^{\mathrm{2}} }{\mathrm{8}{b}}=\frac{\left(\sqrt{{b}}−\sqrt{{a}}\right)^{\mathrm{2}} }{\mathrm{2}}.\frac{\left(\sqrt{{b}}+\sqrt{{a}}\right)^{\mathrm{2}} }{\mathrm{4}{b}}<\frac{\left(\sqrt{{a}}−\sqrt{{b}}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$ $$\frac{\left({b}−{a}\right)^{\mathrm{2}} }{\mathrm{8}{a}}=\frac{\left(\sqrt{{b}}−\sqrt{{a}}\right)^{\mathrm{2}} }{\mathrm{2}}.\frac{\left(\sqrt{{b}}+\sqrt{{a}}\right)^{\mathrm{2}} }{\mathrm{4}{a}}\:>\frac{\left(\sqrt{{a}}−\sqrt{{b}}\right)^{\mathrm{2}} }{\mathrm{2}}\: \\ $$ $$ \\ $$ $${LFYTC} \\ $$ $$ \\ $$

Commented byAr Brandon last updated on 26/Mar/21

Thanks

$$\mathrm{Thanks} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com