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Question Number 6257 by 314159 last updated on 20/Jun/16
Given that a and b are positive real number such that b<4a+1,show that ((2a+b)/(4a+1))<(√(4a^2 +b)) .
Giventhataandbarepositiverealnumbersuchthatb<4a+1,showthat2a+b4a+1<4a2+b.
Answered by Yozzii last updated on 20/Jun/16
((2a+b)/(4a+1))<(√(4a^2 +b)) a,b>0 ⇒(((2a+b)^2 )/((4a+1)^2 ))<4a^2 +b −−−−−−−−−−−−−−−−−−−−−−−−−−− Let φ=(((2a+b)^2 )/((4a+1)^2 ))−4a^2 −b. φ=((4a^2 +4ab+b^2 −(16a^2 +8a+1)(4a^2 +b))/((4a+1)^2 )) φ=((4a^2 +4ab+b^2 −64a^4 −16a^2 b−32a^3 −8ab−4a^2 −b)/((4a+1)^2 )) φ=((−64a^4 −16a^2 b−32a^3 −4ab+b(b−1))/((4a+1)^2 )) Since b<4a+1⇒b−1<4a⇒b(b−1)<4ab ⇒φ<((−64a^4 −16a^2 b−32a^3 −4ab+4ab)/((4a+1)^2 )) φ<((−16a^2 (4a^2 +b+2a))/((4a+1)^2 )) Since a,b>0⇒ 4a^2 +2a+b>0, −16a^2 <0 and (4a+1)^2 >0. Thus, φ<0. ⇒0<(((2a+b)^2 )/((4a+1)^2 ))<4a^2 +b ∴ (√((((2a+b)/(4a+1)))^2 ))<(√(4a^2 +b)) ⇒∣((2a+b)/(4a+1))∣<(√(4a^2 +b)) a,b>0⇒ 2a+b>0 and 4a+1>0. ∴ ((2a+b)/(4a+1))<(√(4a^2 +b)).
2a+b4a+1<4a2+ba,b>0⇒(2a+b)2(4a+1)2<4a2+b−−−−−−−−−−−−−−−−−−−−−−−−−−−Letϕ=(2a+b)2(4a+1)2−4a2−b.ϕ=4a2+4ab+b2−(16a2+8a+1)(4a2+b)(4a+1)2ϕ=4a2+4ab+b2−64a4−16a2b−32a3−8ab−4a2−b(4a+1)2ϕ=−64a4−16a2b−32a3−4ab+b(b−1)(4a+1)2Sinceb<4a+1⇒b−1<4a⇒b(b−1)<4ab⇒ϕ<−64a4−16a2b−32a3−4ab+4ab(4a+1)2ϕ<−16a2(4a2+b+2a)(4a+1)2Sincea,b>0⇒4a2+2a+b>0,−16a2<0and(4a+1)2>0.Thus,ϕ<0.⇒0<(2a+b)2(4a+1)2<4a2+b∴(2a+b4a+1)2<4a2+b⇒∣2a+b4a+1∣<4a2+ba,b>0⇒2a+b>0and4a+1>0.∴2a+b4a+1<4a2+b.

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