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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

$$\mathrm{Thanks} \\ $$

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