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A-2B-3C-4D-are-positive-numbers-forming-a-geometric-series-prov-that-A-3C-B-2D-gt-2-

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Question Number 11956 by Mahmoud A.R last updated on 07/Apr/17
A , 2B ,3C ,4D are positive numbers forming a geometric series prov that : (A + 3C) (B + 2D) > 2
$${A}\:,\:\mathrm{2}{B}\:,\mathrm{3}{C}\:,\mathrm{4}{D}\: \\ $$$${are}\:{positive}\:{numbers}\:{forming}\:{a}\: \\ $$$${geometric}\:{series}\: \\ $$$${prov}\:{that}\:: \\ $$$$\left({A}\:+\:\mathrm{3}{C}\right)\:\left({B}\:+\:\mathrm{2}{D}\right)\:>\:\mathrm{2} \\ $$
Commented by ajfour last updated on 07/Apr/17
if A=(1/(64)), 2B=(1/(32)), 3C=(1/(16)), 4D=(1/8) (A+3C)(B+2D)= (5/(64))×(5/(64)) > 2 !!????
$${if}\:{A}=\frac{\mathrm{1}}{\mathrm{64}},\:\mathrm{2}{B}=\frac{\mathrm{1}}{\mathrm{32}},\:\mathrm{3}{C}=\frac{\mathrm{1}}{\mathrm{16}},\:\mathrm{4}{D}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\left({A}+\mathrm{3}{C}\right)\left({B}+\mathrm{2}{D}\right)=\:\frac{\mathrm{5}}{\mathrm{64}}×\frac{\mathrm{5}}{\mathrm{64}}\:>\:\mathrm{2}\: \\ $$$$!!???? \\ $$
Commented by FilupS last updated on 07/Apr/17
A, B, C, D > 0 Sequence: A, 2B, 3C, 4D ∴2B=nA ⇒ B=((nA)/2) 3C=n(2B)=n^2 A 4D=n(3C)=n^3 A ⇒ 2D=((n^3 A)/(2 )) (A+3C)(B+2D)=(A+n^2 A)(((nA)/2)+((n^3 A)/2)) =A^2 (1+n^2 )((1/2)(n+n^3 )) =(1/2)nA^2 (1+n^2 )(1+n^2 ) =(1/2)nA^2 (1+n^2 )^2 (1+n^2 )^2 >1 A^2 >0 let (1/2)A^2 (1+n^2 )^2 =k, k>0 =nk ????
$${A},\:{B},\:{C},\:{D}\:>\:\mathrm{0} \\ $$$$\: \\ $$$${Sequence}: \\ $$$${A},\:\mathrm{2}{B},\:\mathrm{3}{C},\:\mathrm{4}{D} \\ $$$$\therefore\mathrm{2}{B}={nA}\:\:\Rightarrow\:\:{B}=\frac{{nA}}{\mathrm{2}} \\ $$$$\mathrm{3}{C}={n}\left(\mathrm{2}{B}\right)={n}^{\mathrm{2}} {A} \\ $$$$\mathrm{4}{D}={n}\left(\mathrm{3}{C}\right)={n}^{\mathrm{3}} {A}\:\:\Rightarrow\:\:\mathrm{2}{D}=\frac{{n}^{\mathrm{3}} {A}}{\mathrm{2}\:} \\ $$$$\left({A}+\mathrm{3}{C}\right)\left({B}+\mathrm{2}{D}\right)=\left({A}+{n}^{\mathrm{2}} {A}\right)\left(\frac{{nA}}{\mathrm{2}}+\frac{{n}^{\mathrm{3}} {A}}{\mathrm{2}}\right) \\ $$$$={A}^{\mathrm{2}} \left(\mathrm{1}+{n}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}}{\mathrm{2}}\left({n}+{n}^{\mathrm{3}} \right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{nA}^{\mathrm{2}} \left(\mathrm{1}+{n}^{\mathrm{2}} \right)\left(\mathrm{1}+{n}^{\mathrm{2}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{nA}^{\mathrm{2}} \left(\mathrm{1}+{n}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+{n}^{\mathrm{2}} \right)^{\mathrm{2}} >\mathrm{1} \\ $$$${A}^{\mathrm{2}} >\mathrm{0} \\ $$$$\mathrm{let}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{A}^{\mathrm{2}} \left(\mathrm{1}+{n}^{\mathrm{2}} \right)^{\mathrm{2}} ={k},\:\:\:{k}>\mathrm{0} \\ $$$$={nk} \\ $$$$ \\ $$$$???? \\ $$

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