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