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Question Number 81564 by jagoll last updated on 14/Feb/20
given ((a+b)/c) + ((a+c)/b) +((b+c)/a) = 9 a^2 +b^2 +c^2 = 12 maximum value of a+b+c is
$${given}\: \\ $$$$\frac{{a}+{b}}{{c}}\:+\:\frac{{a}+{c}}{{b}}\:+\frac{{b}+{c}}{{a}}\:=\:\mathrm{9} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\mathrm{12}\: \\ $$$${maximum}\:{value}\:{of}\:{a}+{b}+{c}\:{is} \\ $$
Commented by mr W last updated on 14/Feb/20
a+b+c≤6(√3) ?
$${a}+{b}+{c}\leqslant\mathrm{6}\sqrt{\mathrm{3}}\:? \\ $$
Commented by john santu last updated on 14/Feb/20
let a+b+c = k ((a+b+c−c)/c)+((a+b+c−b)/b)+((a+b+c−a)/a)=9 (k/c)+(k/b)+(k/a)=12⇒ k(((b+c)/(bc))+(1/a))= 12 k(((ab+bc+ac)/(abc)))=12 let ab+bc+ac=m mk = 12abc (i) a^2 +b^2 +c^2 = (a+b+c)^2 −2(ab+ac+bc) ab+ac+bc = ((k^2 −12)/2) (ii)k(((k^2 −12)/(2abc)))=12 ⇒ k^3 −12k=24abc≤24((k/3))^3 27k^3 −324k≤24k^3 3k(k^2 −108)≤0 3k (k+6(√3))(k−6(√3))≤0 for k> 0 ⇒0 < k ≤ 6(√3)
$$\mathrm{let}\:\mathrm{a}+\mathrm{b}+\mathrm{c}\:=\:\mathrm{k} \\ $$$$\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}−\mathrm{c}}{\mathrm{c}}+\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}−\mathrm{b}}{\mathrm{b}}+\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}−\mathrm{a}}{\mathrm{a}}=\mathrm{9} \\ $$$$\frac{\mathrm{k}}{\mathrm{c}}+\frac{\mathrm{k}}{\mathrm{b}}+\frac{\mathrm{k}}{\mathrm{a}}=\mathrm{12}\Rightarrow\: \\ $$$$\mathrm{k}\left(\frac{\mathrm{b}+\mathrm{c}}{\mathrm{bc}}+\frac{\mathrm{1}}{\mathrm{a}}\right)=\:\mathrm{12} \\ $$$$\mathrm{k}\left(\frac{\mathrm{ab}+\mathrm{bc}+\mathrm{ac}}{\mathrm{abc}}\right)=\mathrm{12} \\ $$$$\mathrm{let}\:\mathrm{ab}+\mathrm{bc}+\mathrm{ac}=\mathrm{m} \\ $$$$\mathrm{mk}\:=\:\mathrm{12abc} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \:=\:\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{ab}+\mathrm{ac}+\mathrm{bc}\right) \\ $$$$\mathrm{ab}+\mathrm{ac}+\mathrm{bc}\:=\:\frac{\mathrm{k}^{\mathrm{2}} −\mathrm{12}}{\mathrm{2}} \\ $$$$\left(\mathrm{ii}\right)\mathrm{k}\left(\frac{\mathrm{k}^{\mathrm{2}} −\mathrm{12}}{\mathrm{2abc}}\right)=\mathrm{12}\:\Rightarrow\:\mathrm{k}^{\mathrm{3}} −\mathrm{12k}=\mathrm{24abc}\leqslant\mathrm{24}\left(\frac{\mathrm{k}}{\mathrm{3}}\right)^{\mathrm{3}} \\ $$$$\mathrm{27k}^{\mathrm{3}} −\mathrm{324k}\leqslant\mathrm{24k}^{\mathrm{3}} \\ $$$$\mathrm{3k}\left(\mathrm{k}^{\mathrm{2}} −\mathrm{108}\right)\leqslant\mathrm{0}\: \\ $$$$\mathrm{3k}\:\left(\mathrm{k}+\mathrm{6}\sqrt{\mathrm{3}}\right)\left(\mathrm{k}−\mathrm{6}\sqrt{\mathrm{3}}\right)\leqslant\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{k}>\:\mathrm{0}\:\Rightarrow\mathrm{0}\:<\:\mathrm{k}\:\leqslant\:\mathrm{6}\sqrt{\mathrm{3}}\: \\ $$
Commented by mr W last updated on 14/Feb/20
you need only a little step to the final solution: k^3 −12k=24abc≤24(((a+b+c)/3))^3 =((24k^3 )/(27)) ⇒(k^3 /9)−12k≤0 ⇒k≤(√(9×12))=6(√3)
$${you}\:{need}\:{only}\:{a}\:{little}\:{step}\:{to}\:{the}\:{final} \\ $$$${solution}: \\ $$$$\mathrm{k}^{\mathrm{3}} −\mathrm{12k}=\mathrm{24abc}\leqslant\mathrm{24}\left(\frac{{a}+{b}+{c}}{\mathrm{3}}\right)^{\mathrm{3}} =\frac{\mathrm{24}{k}^{\mathrm{3}} }{\mathrm{27}} \\ $$$$\Rightarrow\frac{{k}^{\mathrm{3}} }{\mathrm{9}}−\mathrm{12k}\leqslant\mathrm{0} \\ $$$$\Rightarrow{k}\leqslant\sqrt{\mathrm{9}×\mathrm{12}}=\mathrm{6}\sqrt{\mathrm{3}} \\ $$
Commented by mr W last updated on 14/Feb/20
please discuss: can a+b+c reach this maximum 6(√3)? i mean, is a+b+c<6(√3) or a+b+c≤6(√3) correct?
$${please}\:{discuss}: \\ $$$${can}\:{a}+{b}+{c}\:{reach}\:{this}\:{maximum}\:\mathrm{6}\sqrt{\mathrm{3}}? \\ $$$${i}\:{mean},\:{is}\:{a}+{b}+{c}<\mathrm{6}\sqrt{\mathrm{3}}\:{or} \\ $$$${a}+{b}+{c}\leqslant\mathrm{6}\sqrt{\mathrm{3}}\:{correct}? \\ $$
Commented by john santu last updated on 14/Feb/20
we use AM sir? yes i agree with you
$$\mathrm{we}\:\mathrm{use}\:\mathrm{AM}\:\mathrm{sir}?\:\mathrm{yes}\:\mathrm{i}\:\mathrm{agree}\:\mathrm{with}\:\mathrm{you} \\ $$
Commented by mr W last updated on 14/Feb/20
my question is if the “=” sign in a+b+c≤6(√3) is valid. i think it isn′t. i.e. correct is: a+b+c<6(√3). what′s your opinion?
$${my}\:{question}\:{is}\:{if}\:{the}\:“=''\:{sign}\:{in} \\ $$$${a}+{b}+{c}\leqslant\mathrm{6}\sqrt{\mathrm{3}}\:{is}\:{valid}.\:\:{i}\:{think}\:{it}\:{isn}'{t}. \\ $$$${i}.{e}.\:{correct}\:{is}: \\ $$$${a}+{b}+{c}<\mathrm{6}\sqrt{\mathrm{3}}. \\ $$$${what}'{s}\:{your}\:{opinion}? \\ $$
Commented by jagoll last updated on 14/Feb/20
i not find this solution sir.
$${i}\:{not}\:{find}\:{this}\:{solution}\:{sir}. \\ $$
Commented by mr W last updated on 14/Feb/20
you mean this solution is wrong or you mean you don′t know the correct solution?
$${you}\:{mean}\:{this}\:{solution}\:{is}\:{wrong}\:{or} \\ $$$${you}\:{mean}\:{you}\:{don}'{t}\:{know}\:{the}\:{correct} \\ $$$${solution}? \\ $$
Commented by jagoll last updated on 14/Feb/20
i think this solution is correct. but i don′t know the solution does it match in the book? answer in the book were not included
$${i}\:{think}\:{this}\:{solution}\:{is}\:{correct}. \\ $$$${but}\:{i}\:{don}'{t}\:{know}\:{the}\:{solution}\:{does} \\ $$$${it}\:{match}\:{in}\:{the}\:{book}?\:{answer} \\ $$$${in}\:{the}\:{book}\:{were}\:{not}\:{included} \\ $$
Commented by mr W last updated on 14/Feb/20
k^3 −12k=24abc≤ ((k/3))^3 is wrong. correct is: k^3 −12k=24abc≤ 24((k/3))^3 . this is for k>0. if k<0, then k^3 −12k=24abc≥−24((k/3))^3 .
$$\mathrm{k}^{\mathrm{3}} −\mathrm{12k}=\mathrm{24abc}\leqslant\:\left(\frac{\mathrm{k}}{\mathrm{3}}\right)^{\mathrm{3}} \:{is}\:{wrong}. \\ $$$${correct}\:{is}: \\ $$$$\mathrm{k}^{\mathrm{3}} −\mathrm{12k}=\mathrm{24abc}\leqslant\:\mathrm{24}\left(\frac{\mathrm{k}}{\mathrm{3}}\right)^{\mathrm{3}} . \\ $$$${this}\:{is}\:{for}\:{k}>\mathrm{0}. \\ $$$$ \\ $$$${if}\:{k}<\mathrm{0},\:{then} \\ $$$$\mathrm{k}^{\mathrm{3}} −\mathrm{12k}=\mathrm{24abc}\geqslant−\mathrm{24}\left(\frac{\mathrm{k}}{\mathrm{3}}\right)^{\mathrm{3}} . \\ $$
Commented by john santu last updated on 14/Feb/20
oo yes it my typo
$$\mathrm{oo}\:\mathrm{yes}\:\mathrm{it}\:\mathrm{my}\:\mathrm{typo} \\ $$

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