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Question Number 39868 by math2018 last updated on 12/Jul/18
a,b,c>0,if ((8a^2 )/(a^2 +9))=b, ((10b^2 )/(b^2 +16))=c, ((6c^2 )/(c^2 +25))=a, then,what is the minimum value of a+b+c?
$${a},{b},{c}>\mathrm{0},{if} \\ $$$$\frac{\mathrm{8}{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{9}}={b},\:\:\frac{\mathrm{10}{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} +\mathrm{16}}={c},\:\:\frac{\mathrm{6}{c}^{\mathrm{2}} }{{c}^{\mathrm{2}} +\mathrm{25}}={a}, \\ $$$${then},{what}\:{is}\:{the}\:{minimum}\:{value}\:{of}\:\:{a}+{b}+{c}? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jul/18
b=(8/(1+(9/a^2 )))=(8/((1−(3/a))^2 +(6/a))) if a=3 then b=4 c=((10)/((1−(4/b))^2 +(8/b))) if b=4 c=5 a=(6/((1−(5/c))^2 +((10)/c))) if c=5 a=3 so a=3 b=4 c=5 satisfy three eqn a+b+c=12 pls check...whether any direct method to solve i do not know...
$${b}=\frac{\mathrm{8}}{\mathrm{1}+\frac{\mathrm{9}}{{a}^{\mathrm{2}} }}=\frac{\mathrm{8}}{\left(\mathrm{1}−\frac{\mathrm{3}}{{a}}\right)^{\mathrm{2}} +\frac{\mathrm{6}}{{a}}} \\ $$$${if}\:\:{a}=\mathrm{3}\:\:{then}\:\:\:{b}=\mathrm{4} \\ $$$${c}=\frac{\mathrm{10}}{\left(\mathrm{1}−\frac{\mathrm{4}}{{b}}\right)^{\mathrm{2}} +\frac{\mathrm{8}}{{b}}}\:\:{if}\:{b}=\mathrm{4}\:\:\:\:{c}=\mathrm{5} \\ $$$${a}=\frac{\mathrm{6}}{\left(\mathrm{1}−\frac{\mathrm{5}}{{c}}\right)^{\mathrm{2}} +\frac{\mathrm{10}}{{c}}}\:\:{if}\:{c}=\mathrm{5}\:\:\:{a}=\mathrm{3} \\ $$$${so}\:{a}=\mathrm{3} \\ $$$${b}=\mathrm{4} \\ $$$${c}=\mathrm{5}\:{satisfy}\:{three}\:{eqn} \\ $$$${a}+{b}+{c}=\mathrm{12} \\ $$$${pls}\:{check}…{whether}\:{any}\:{direct}\:{method}\:{to}\:{solve} \\ $$$${i}\:{do}\:{not}\:{know}… \\ $$$$ \\ $$
Commented by math2018 last updated on 13/Jul/18
Thank you for your work!
$${Thank}\:{you}\:{for}\:{your}\:{work}! \\ $$
Answered by ajfour last updated on 12/Jul/18
let a=3tan 𝛂, b=4tan 𝛃, c=5tan 𝛄 Then ((8×9tan^2 α)/(9sec^2 α)) =4tan β ⇒ 2sin^2 α = tan β ....(i) ((10×16tan^2 β)/(16sec^2 β)) =5tan γ ⇒ 2sin^2 β = tan γ ....(ii) ((6×25tan^2 γ)/(25sec^2 γ)) = 3tan α ⇒ 2sin^2 γ = tan α ....(iii) (i)×(ii)×(iii) gives 8sin^2 α sin^2 β sin^2 γ =tan α tan β tan γ or sin 2α sin 2β sin 2γ = 1 ⇒ α = β = γ = (π/4) a+b+c = 3tan α+4tan β+5tan γ = (3+4+5)tan (π/4) a+b+c = 12 . (why the minimum? )!
$${let}\:\boldsymbol{{a}}=\mathrm{3tan}\:\boldsymbol{\alpha},\:\boldsymbol{{b}}=\mathrm{4tan}\:\boldsymbol{\beta},\:\boldsymbol{{c}}=\mathrm{5tan}\:\boldsymbol{\gamma} \\ $$$${Then} \\ $$$$\frac{\mathrm{8}×\mathrm{9tan}\:^{\mathrm{2}} \alpha}{\mathrm{9sec}\:^{\mathrm{2}} \alpha}\:=\mathrm{4tan}\:\beta \\ $$$$\Rightarrow\:\:\mathrm{2sin}\:^{\mathrm{2}} \alpha\:=\:\mathrm{tan}\:\beta\:\:\:\:\:\:\:….\left({i}\right) \\ $$$$\frac{\mathrm{10}×\mathrm{16tan}\:^{\mathrm{2}} \beta}{\mathrm{16sec}\:^{\mathrm{2}} \beta}\:=\mathrm{5tan}\:\gamma \\ $$$$\Rightarrow\:\:\mathrm{2sin}\:^{\mathrm{2}} \beta\:=\:\mathrm{tan}\:\gamma\:\:\:\:\:\:….\left({ii}\right) \\ $$$$\frac{\mathrm{6}×\mathrm{25tan}\:^{\mathrm{2}} \gamma}{\mathrm{25sec}\:^{\mathrm{2}} \gamma}\:=\:\mathrm{3tan}\:\alpha \\ $$$$\Rightarrow\:\:\mathrm{2sin}\:^{\mathrm{2}} \gamma\:=\:\mathrm{tan}\:\alpha\:\:\:\:\:\:\:\:….\left({iii}\right) \\ $$$$\left({i}\right)×\left({ii}\right)×\left({iii}\right)\:\:\:{gives} \\ $$$$\mathrm{8sin}\:^{\mathrm{2}} \alpha\:\mathrm{sin}\:^{\mathrm{2}} \beta\:\mathrm{sin}\:^{\mathrm{2}} \gamma\:=\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma \\ $$$${or}\:\:\:\:\mathrm{sin}\:\mathrm{2}\alpha\:\mathrm{sin}\:\mathrm{2}\beta\:\mathrm{sin}\:\mathrm{2}\gamma\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\:\alpha\:=\:\beta\:=\:\gamma\:=\:\frac{\pi}{\mathrm{4}} \\ $$$${a}+{b}+{c}\:=\:\mathrm{3tan}\:\alpha+\mathrm{4tan}\:\beta+\mathrm{5tan}\:\gamma \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{3}+\mathrm{4}+\mathrm{5}\right)\mathrm{tan}\:\frac{\pi}{\mathrm{4}} \\ $$$${a}+{b}+{c}\:=\:\mathrm{12}\:. \\ $$$$\left({why}\:{the}\:{minimum}?\:\right)! \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18
yes i forgot it ...thank you for correction
$${yes}\:\:{i}\:{forgot}\:{it}\:…{thank}\:{you}\:{for}\:{correction} \\ $$
Commented by math2018 last updated on 13/Jul/18
∵a>0,b>0,c>0 ∴a+b+c≠0 Thank you Sir.
$$\because{a}>\mathrm{0},{b}>\mathrm{0},{c}>\mathrm{0} \\ $$$$\therefore{a}+{b}+{c}\neq\mathrm{0} \\ $$$${Thank}\:{you}\:{Sir}. \\ $$
Commented by math2018 last updated on 13/Jul/18
a=((6c^2 )/(c^2 +25))≤((6c^2 )/(2(√(25c^2 ))))=((3c)/5) b=((8a^2 )/(a^2 +9))≤((8a^2 )/(2(√(9a^2 ))))=((4a)/3) c=((10b^2 )/(b^2 +16))≤((10b^2 )/(2(√(16b^2 ))))=((5b)/4) now we can get the minimum value of a+b+c Thank you very much!
$${a}=\frac{\mathrm{6}{c}^{\mathrm{2}} }{{c}^{\mathrm{2}} +\mathrm{25}}\leqslant\frac{\mathrm{6}{c}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{25}{c}^{\mathrm{2}} }}=\frac{\mathrm{3}{c}}{\mathrm{5}} \\ $$$${b}=\frac{\mathrm{8}{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{9}}\leqslant\frac{\mathrm{8}{a}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{9}{a}^{\mathrm{2}} }}=\frac{\mathrm{4}{a}}{\mathrm{3}} \\ $$$${c}=\frac{\mathrm{10}{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} +\mathrm{16}}\leqslant\frac{\mathrm{10}{b}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{16}{b}^{\mathrm{2}} }}=\frac{\mathrm{5}{b}}{\mathrm{4}} \\ $$$${now}\:{we}\:{can}\:{get}\:{the}\:{minimum}\:{value}\:{of}\:{a}+{b}+{c} \\ $$$${Thank}\:{you}\:{very}\:{much}! \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18
(a/3)=(b/4)=(c/5)=k a=3k b=4k c=5k a=((6c^2 )/(c^2 +25))=((6×25k^2 )/(25k^2 +25)) 3k=((6k^2 )/(k^2 +1)) 3k(k^2 +1)−6k^2 =0 3k(k^2 +1−2k)=0 so k=0 and k=1 a+b+c =3k+4k+5k=12k so when k=0 a+b+c=0 when k=1 a+b+c=12 hence min value is zero...pls check...
$$\frac{{a}}{\mathrm{3}}=\frac{{b}}{\mathrm{4}}=\frac{{c}}{\mathrm{5}}={k} \\ $$$${a}=\mathrm{3}{k}\:\:{b}=\mathrm{4}{k}\:\:{c}=\mathrm{5}{k}\:\: \\ $$$${a}=\frac{\mathrm{6}{c}^{\mathrm{2}} }{{c}^{\mathrm{2}} +\mathrm{25}}=\frac{\mathrm{6}×\mathrm{25}{k}^{\mathrm{2}} }{\mathrm{25}{k}^{\mathrm{2}} +\mathrm{25}} \\ $$$$\mathrm{3}{k}=\frac{\mathrm{6}{k}^{\mathrm{2}} }{{k}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{3}{k}\left({k}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{6}{k}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{3}{k}\left({k}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{k}\right)=\mathrm{0} \\ $$$${so}\:{k}=\mathrm{0}\:{and}\:{k}=\mathrm{1} \\ $$$${a}+{b}+{c} \\ $$$$=\mathrm{3}{k}+\mathrm{4}{k}+\mathrm{5}{k}=\mathrm{12}{k} \\ $$$${so}\:{when}\:{k}=\mathrm{0}\:\:\:{a}+{b}+{c}=\mathrm{0} \\ $$$${when}\:{k}=\mathrm{1}\:\:{a}+{b}+{c}=\mathrm{12} \\ $$$${hence}\:{min}\:{value}\:{is}\:{zero}…{pls}\:{check}… \\ $$$$ \\ $$

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