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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)) .

$${Given}\:{that}\:{a}\:{and}\:{b}\:{are}\:{positive}\:{real}\:{number} \\ $$ $${such}\:{that}\:{b}<\mathrm{4}{a}+\mathrm{1},{show}\:{that}\:\frac{\mathrm{2}{a}+{b}}{\mathrm{4}{a}+\mathrm{1}}<\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{b}}\:. \\ $$

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)).

$$\frac{\mathrm{2}{a}+{b}}{\mathrm{4}{a}+\mathrm{1}}<\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{b}}\:\:\:\:\:\:{a},{b}>\mathrm{0} \\ $$ $$\Rightarrow\frac{\left(\mathrm{2}{a}+{b}\right)^{\mathrm{2}} }{\left(\mathrm{4}{a}+\mathrm{1}\right)^{\mathrm{2}} }<\mathrm{4}{a}^{\mathrm{2}} +{b} \\ $$ $$−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$ $${Let}\:\phi=\frac{\left(\mathrm{2}{a}+{b}\right)^{\mathrm{2}} }{\left(\mathrm{4}{a}+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{4}{a}^{\mathrm{2}} −{b}. \\ $$ $$\phi=\frac{\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{ab}+{b}^{\mathrm{2}} −\left(\mathrm{16}{a}^{\mathrm{2}} +\mathrm{8}{a}+\mathrm{1}\right)\left(\mathrm{4}{a}^{\mathrm{2}} +{b}\right)}{\left(\mathrm{4}{a}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ $$\phi=\frac{\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{ab}+{b}^{\mathrm{2}} −\mathrm{64}{a}^{\mathrm{4}} −\mathrm{16}{a}^{\mathrm{2}} {b}−\mathrm{32}{a}^{\mathrm{3}} −\mathrm{8}{ab}−\mathrm{4}{a}^{\mathrm{2}} −{b}}{\left(\mathrm{4}{a}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ $$\phi=\frac{−\mathrm{64}{a}^{\mathrm{4}} −\mathrm{16}{a}^{\mathrm{2}} {b}−\mathrm{32}{a}^{\mathrm{3}} −\mathrm{4}{ab}+{b}\left({b}−\mathrm{1}\right)}{\left(\mathrm{4}{a}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ $${Since}\:{b}<\mathrm{4}{a}+\mathrm{1}\Rightarrow{b}−\mathrm{1}<\mathrm{4}{a}\Rightarrow{b}\left({b}−\mathrm{1}\right)<\mathrm{4}{ab} \\ $$ $$\Rightarrow\phi<\frac{−\mathrm{64}{a}^{\mathrm{4}} −\mathrm{16}{a}^{\mathrm{2}} {b}−\mathrm{32}{a}^{\mathrm{3}} −\mathrm{4}{ab}+\mathrm{4}{ab}}{\left(\mathrm{4}{a}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ $$\phi<\frac{−\mathrm{16}{a}^{\mathrm{2}} \left(\mathrm{4}{a}^{\mathrm{2}} +{b}+\mathrm{2}{a}\right)}{\left(\mathrm{4}{a}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ $${Since}\:{a},{b}>\mathrm{0}\Rightarrow\:\mathrm{4}{a}^{\mathrm{2}} +\mathrm{2}{a}+{b}>\mathrm{0},\:−\mathrm{16}{a}^{\mathrm{2}} <\mathrm{0} \\ $$ $${and}\:\left(\mathrm{4}{a}+\mathrm{1}\right)^{\mathrm{2}} >\mathrm{0}.\:{Thus},\:\phi<\mathrm{0}. \\ $$ $$\Rightarrow\mathrm{0}<\frac{\left(\mathrm{2}{a}+{b}\right)^{\mathrm{2}} }{\left(\mathrm{4}{a}+\mathrm{1}\right)^{\mathrm{2}} }<\mathrm{4}{a}^{\mathrm{2}} +{b} \\ $$ $$\therefore\:\sqrt{\left(\frac{\mathrm{2}{a}+{b}}{\mathrm{4}{a}+\mathrm{1}}\right)^{\mathrm{2}} }<\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{b}} \\ $$ $$\Rightarrow\mid\frac{\mathrm{2}{a}+{b}}{\mathrm{4}{a}+\mathrm{1}}\mid<\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{b}} \\ $$ $${a},{b}>\mathrm{0}\Rightarrow\:\mathrm{2}{a}+{b}>\mathrm{0}\:{and}\:\mathrm{4}{a}+\mathrm{1}>\mathrm{0}. \\ $$ $$\therefore\:\frac{\mathrm{2}{a}+{b}}{\mathrm{4}{a}+\mathrm{1}}<\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{b}}. \\ $$

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