Question-204756

Question Number 204756 by York12 last updated on 27/Feb/24

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Commented by Rasheed.Sindhi last updated on 27/Feb/24

Wrong! Given condition satisfied by: a=0,b=0,c=1 (0+0+1)^2 =0^2 +0^2 +1^2 But the result to be proved is false. you can see: 0=1

$$\mathrm{Wrong}! \\ $$$${Given}\:{condition}\:{satisfied}\:{by}: \\ $$$$\:\:{a}=\mathrm{0},{b}=\mathrm{0},{c}=\mathrm{1} \\ $$$$\left(\mathrm{0}+\mathrm{0}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}^{\mathrm{2}} +\mathrm{0}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} \\ $$$${But}\:{the}\:{result}\:{to}\:{be}\:{proved}\:{is}\:{false}. \\ $$$${you}\:{can}\:{see}:\:\mathrm{0}=\mathrm{1} \\ $$

Commented by York12 last updated on 27/Feb/24

Commented by York12 last updated on 27/Feb/24

Commented by York12 last updated on 27/Feb/24

$$\mathrm{I}\:\mathrm{have}\:\mathrm{proved}\:\mathrm{it}\:\mathrm{though}\:!! \\ $$

Commented by Rasheed.Sindhi last updated on 27/Feb/24

$$\mathcal{T}{hen}\:{there}\:{must}\:{be}\:{flaw}\:{in}\:{your} \\ $$$${proof}! \\ $$

Commented by York12 last updated on 27/Feb/24

maybe , but can you see where the mistake is

$$\mathrm{maybe}\:,\:\mathrm{but}\:\mathrm{can}\:\mathrm{you}\:\mathrm{see}\:\mathrm{where}\:\mathrm{the}\:\mathrm{mistake}\:\mathrm{is} \\ $$

Commented by York12 last updated on 27/Feb/24

sir there is no such a triple cause then ((ac)/(b^2 +ac))=(0/0) !

$$\mathrm{sir}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{such}\:\mathrm{a}\:\mathrm{triple}\:\mathrm{cause}\:\mathrm{then}\:\frac{\mathrm{ac}}{\mathrm{b}^{\mathrm{2}} +\mathrm{ac}}=\frac{\mathrm{0}}{\mathrm{0}}\:! \\ $$

Commented by Rasheed.Sindhi last updated on 27/Feb/24

$${Anyway}\:{this}\:{triple}\:{disproves}\:{the} \\ $$$${result}! \\ $$

Commented by Rasheed.Sindhi last updated on 27/Feb/24

$${BTW}\:{I}\:{think},\:{it}'{s}\:{not}\:{good}\:{practice} \\ $$$${to}\:{post}\:{question}\:{as}\:{a}\:{comment}\:{of} \\ $$$${empty}\:{question}-{post}! \\ $$$$ \\ $$

Commented by York12 last updated on 27/Feb/24

no I am saying you are wrong Σ((ab)/(c^2 +2ab)) ≠ 0 at (a,b,c)=(0,0,1)_! but it is undefined

$$\mathrm{no}\:\mathrm{I}\:\mathrm{am}\:\mathrm{saying}\:\mathrm{you}\:\mathrm{are}\:\mathrm{wrong} \\ $$$$\Sigma\frac{\mathrm{ab}}{\mathrm{c}^{\mathrm{2}} +\mathrm{2ab}}\:\neq\:\mathrm{0}\:\mathrm{at}\:\left({a},{b},{c}\right)=\left(\mathrm{0},\mathrm{0},\mathrm{1}\right)_{!} \:\mathrm{but}\:\mathrm{it}\:\mathrm{is}\:\mathrm{undefined} \\ $$

Commented by York12 last updated on 27/Feb/24

$$\mathrm{my}\:\mathrm{phone}\:\mathrm{is}\:\mathrm{weak}\:\mathrm{I}\:\mathrm{can}\:\mathrm{not}\:\mathrm{post}\:\mathrm{at}\:\mathrm{once} \\ $$

Commented by A5T last updated on 27/Feb/24

I guess the question doesn′t actually claim that a^2 +b^2 +c^2 =(a+b+c)^2 ; it′s just given as a condition, sort of “given that a^2 +b^2 +c^2 =(a+b+c)^2 , prove that...”. a^2 +b^2 +c^2 =(a+b+c)^2 ⇒ab+bc+ca=0 which was used in the solution.

$${I}\:{guess}\:{the}\:{question}\:{doesn}'{t}\:{actually}\:{claim}\:{that} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\left({a}+{b}+{c}\right)^{\mathrm{2}} ;\:{it}'{s}\:{just}\:{given}\:{as}\:{a}\: \\ $$$${condition},\:{sort}\:{of}\:“{given}\:{that}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\left({a}+{b}+{c}\right)^{\mathrm{2}} , \\ $$$${prove}\:{that}…''.\: \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\left({a}+{b}+{c}\right)^{\mathrm{2}} \Rightarrow{ab}+{bc}+{ca}=\mathrm{0}\:{which}\:{was} \\ $$$${used}\:{in}\:{the}\:{solution}. \\ $$

Commented by York12 last updated on 27/Feb/24

how are those different (a+b+c)^2 =a^2 +b^2 +c^2 ⇔ab+bc+ac=0

$$\mathrm{how}\:\mathrm{are}\:\mathrm{those}\:\mathrm{different}\: \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \Leftrightarrow{ab}+{bc}+{ac}=\mathrm{0} \\ $$

Commented by A5T last updated on 27/Feb/24

I never claimed they were different, they imply each other. I was just pointing out what the (a+b+c)^2 =a^2 +b^2 +c^2 condition implied.

$${I}\:{never}\:{claimed}\:{they}\:{were}\:{different},\:{they}\:{imply} \\ $$$${each}\:{other}.\:{I}\:{was}\:{just}\:{pointing}\:{out}\:{what}\:{the} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:{condition}\:{implied}. \\ $$

Commented by A5T last updated on 27/Feb/24

If the proof is correct, then Σ((bc)/(a^2 +2bc))=1 when all terms are defined, but the case of (0,0,1), not all terms are defined. Since the proof assumed all the terms are defined, that additional condition could have been given.

$${If}\:{the}\:{proof}\:{is}\:{correct},\:{then}\:\Sigma\frac{{bc}}{{a}^{\mathrm{2}} +\mathrm{2}{bc}}=\mathrm{1}\:{when} \\ $$$${all}\:{terms}\:{are}\:{defined},\:{but}\:{the}\:{case}\:{of}\:\left(\mathrm{0},\mathrm{0},\mathrm{1}\right), \\ $$$${not}\:{all}\:{terms}\:{are}\:{defined}.\:{Since}\:{the}\:{proof}\: \\ $$$${assumed}\:{all}\:{the}\:{terms}\:{are}\:{defined},\:{that} \\ $$$${additional}\:{condition}\:{could}\:{have}\:{been}\:{given}. \\ $$

Commented by Rasheed.Sindhi last updated on 27/Feb/24

$${yes}\:{sir},\:{additional}\:{condition}\:{must} \\ $$$${be}\:{given}. \\ $$$${BTW}\:{do}\:{you}\:{think}\:{the}\:{proof}\:{correct}? \\ $$

Commented by York12 last updated on 27/Feb/24

no (0,0,1) is not defined ((bc)/(a^2 +2bc))=((0×c)/(0^2 +2×0×c))=(0/0) →(not defined) ((ac)/(b^2 +2ac))=((0×c)/(0^2 +2×0×0))=(0/0)→(not defined) ((ab)/(c^2 +2ab))=(0/(1+0))=0

$$\mathrm{no}\:\left(\mathrm{0},\mathrm{0},\mathrm{1}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined} \\ $$$$ \\ $$$$\frac{\mathrm{bc}}{\mathrm{a}^{\mathrm{2}} +\mathrm{2bc}}=\frac{\mathrm{0}×{c}}{\mathrm{0}^{\mathrm{2}} +\mathrm{2}×\mathrm{0}×{c}}=\frac{\mathrm{0}}{\mathrm{0}}\:\rightarrow\left(\mathrm{not}\:\mathrm{defined}\right) \\ $$$$\frac{{ac}}{{b}^{\mathrm{2}} +\mathrm{2}{ac}}=\frac{\mathrm{0}×{c}}{\mathrm{0}^{\mathrm{2}} +\mathrm{2}×\mathrm{0}×\mathrm{0}}=\frac{\mathrm{0}}{\mathrm{0}}\rightarrow\left(\mathrm{not}\:\mathrm{defined}\right) \\ $$$$\frac{{ab}}{{c}^{\mathrm{2}} +\mathrm{2}{ab}}=\frac{\mathrm{0}}{\mathrm{1}+\mathrm{0}}=\mathrm{0} \\ $$

Commented by York12 last updated on 27/Feb/24

$$\mathrm{No}\:\mathrm{concentrate}\:\mathrm{well} \\ $$

Commented by A5T last updated on 27/Feb/24

Did I claim (0,0,1) was defined? I guess you are the one who should concentrate well.

$${Did}\:{I}\:{claim}\:\left(\mathrm{0},\mathrm{0},\mathrm{1}\right)\:{was}\:{defined}?\:{I}\:{guess}\:{you} \\ $$$${are}\:{the}\:{one}\:{who}\:{should}\:{concentrate}\:{well}. \\ $$

Commented by A5T last updated on 27/Feb/24

$${The}\:{statement}\:{is}\:{correct}\:{provided}\:{all}\:{the}\:{terms} \\ $$$${are}\:{defined}.\:{I}\:{have}\:{a}\:{different}\:{proof}\:{below} \\ $$$${by}\:{expansion}\:{and}\:{simplification}. \\ $$

Commented by Frix last updated on 27/Feb/24

It′s claimed a, b, c ∈R. Thus it′s not true in some cases. It must be stated when exactly it′s true otherwise the proof is incomplete.

$$\mathrm{It}'\mathrm{s}\:\mathrm{claimed}\:{a},\:{b},\:{c}\:\in\mathbb{R}.\:\mathrm{Thus}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{true}\:\mathrm{in} \\ $$$$\mathrm{some}\:\mathrm{cases}.\:\mathrm{It}\:\mathrm{must}\:\mathrm{be}\:\mathrm{stated}\:\mathrm{when}\:\mathrm{exactly} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{true}\:\mathrm{otherwise}\:\mathrm{the}\:\mathrm{proof}\:\mathrm{is}\:\mathrm{incomplete}. \\ $$

Commented by A5T last updated on 27/Feb/24

The statement is true when none of a^2 +2bc, b^2 +2ac,c^2 +2ab is 0 or otherwise when all of ((bc)/(a^2 +2bc));((ac)/(b^2 +2ac));((ab)/(c^2 +2ab)) are defined.

$${The}\:{statement}\:{is}\:{true}\:{when}\:{none}\:{of}\:{a}^{\mathrm{2}} +\mathrm{2}{bc}, \\ $$$${b}^{\mathrm{2}} +\mathrm{2}{ac},{c}^{\mathrm{2}} +\mathrm{2}{ab}\:{is}\:\mathrm{0}\:{or}\:{otherwise}\:{when}\:{all}\:{of} \\ $$$$\frac{{bc}}{{a}^{\mathrm{2}} +\mathrm{2}{bc}};\frac{{ac}}{{b}^{\mathrm{2}} +\mathrm{2}{ac}};\frac{{ab}}{{c}^{\mathrm{2}} +\mathrm{2}{ab}}\:{are}\:{defined}. \\ $$

Commented by York12 last updated on 27/Feb/24

$$\mathrm{hmmm}\:\mathrm{okay} \\ $$

Answered by A5T last updated on 27/Feb/24

=((Σ(b^3 c^3 +2ab^4 c+2abc^4 +4a^2 b^2 c^2 ))/(a^2 b^2 c^2 +2a^3 b^3 +2a^3 c^3 +2b^3 c^3 +4abc(a^3 +b^3 +c^3 )+8Πa^2 )) =((15a^2 b^2 c^2 +4abc(a^3 +b^3 +c^3 ))/(15a^2 b^2 c^2 +4abc(a^3 +b^3 +c^3 )))=1 Details: Σ[bc(b^2 +2ac)(c^2 +2ab)] =Σ(b^3 c^3 +2ab^4 c+2abc^4 +4a^2 b^2 c^2 ) =Σa^3 b^3 +Σ2abc(a^3 +b^3 )+12a^2 b^2 c^2 =a^3 b^3 +b^3 c^3 +c^3 a^3 +4abc(a^3 +b^3 +c^3 )+12a^2 b^2 c^2 (ab+bc)^3 +(ac)^3 −3ab^2 c(ab+bc)+4abc(Σa^3 )+12Πa^2 =−3ab^2 c(ab+bc)−3ab^2 c(ca)+3a^2 b^2 c^2 +12Πa^2 +4abc(Σa^3 ) =15a^2 b^2 c^2 +4abc(Σa^3 ) (a^2 +2bc)(b^2 +2ac)(c^2 +2ab) =(a^2 +2bc)(b^2 c^2 +2ab^3 +2ac^3 +4a^2 bc) =a^2 b^2 c^2 +2a^3 b^3 +2a^3 c^3 +4a^3 bc+2b^3 c^3 +4ab^4 c +4abc^4 +8a^2 b^2 c^2 =7a^2 b^2 c^2 +4abc(a^3 +b^3 +c^3 )+8a^2 b^2 c^2

$$=\frac{\Sigma\left({b}^{\mathrm{3}} {c}^{\mathrm{3}} +\mathrm{2}{ab}^{\mathrm{4}} {c}+\mathrm{2}{abc}^{\mathrm{4}} +\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{3}} {b}^{\mathrm{3}} +\mathrm{2}{a}^{\mathrm{3}} {c}^{\mathrm{3}} +\mathrm{2}{b}^{\mathrm{3}} {c}^{\mathrm{3}} +\mathrm{4}{abc}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)+\mathrm{8}\Pi{a}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{15}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{4}{abc}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)}{\mathrm{15}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{4}{abc}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)}=\mathrm{1} \\ $$$$ \\ $$$$\underline{{Details}:} \\ $$$$\Sigma\left[{bc}\left({b}^{\mathrm{2}} +\mathrm{2}{ac}\right)\left({c}^{\mathrm{2}} +\mathrm{2}{ab}\right)\right] \\ $$$$=\Sigma\left({b}^{\mathrm{3}} {c}^{\mathrm{3}} +\mathrm{2}{ab}^{\mathrm{4}} {c}+\mathrm{2}{abc}^{\mathrm{4}} +\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \right) \\ $$$$=\Sigma{a}^{\mathrm{3}} {b}^{\mathrm{3}} +\Sigma\mathrm{2}{abc}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)+\mathrm{12}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \\ $$$$={a}^{\mathrm{3}} {b}^{\mathrm{3}} +{b}^{\mathrm{3}} {c}^{\mathrm{3}} +{c}^{\mathrm{3}} {a}^{\mathrm{3}} +\mathrm{4}{abc}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)+\mathrm{12}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \\ $$$$\left({ab}+{bc}\right)^{\mathrm{3}} +\left({ac}\right)^{\mathrm{3}} −\mathrm{3}{ab}^{\mathrm{2}} {c}\left({ab}+{bc}\right)+\mathrm{4}{abc}\left(\Sigma{a}^{\mathrm{3}} \right)+\mathrm{12}\Pi{a}^{\mathrm{2}} \\ $$$$=−\mathrm{3}{ab}^{\mathrm{2}} {c}\left({ab}+{bc}\right)−\mathrm{3}{ab}^{\mathrm{2}} {c}\left({ca}\right)+\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{12}\Pi{a}^{\mathrm{2}} +\mathrm{4}{abc}\left(\Sigma{a}^{\mathrm{3}} \right) \\ $$$$=\mathrm{15}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{4}{abc}\left(\Sigma{a}^{\mathrm{3}} \right) \\ $$$$ \\ $$$$\left({a}^{\mathrm{2}} +\mathrm{2}{bc}\right)\left({b}^{\mathrm{2}} +\mathrm{2}{ac}\right)\left({c}^{\mathrm{2}} +\mathrm{2}{ab}\right) \\ $$$$=\left({a}^{\mathrm{2}} +\mathrm{2}{bc}\right)\left({b}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{2}{ab}^{\mathrm{3}} +\mathrm{2}{ac}^{\mathrm{3}} +\mathrm{4}{a}^{\mathrm{2}} {bc}\right) \\ $$$$={a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{3}} {b}^{\mathrm{3}} +\mathrm{2}{a}^{\mathrm{3}} {c}^{\mathrm{3}} +\mathrm{4}{a}^{\mathrm{3}} {bc}+\mathrm{2}{b}^{\mathrm{3}} {c}^{\mathrm{3}} +\mathrm{4}{ab}^{\mathrm{4}} {c} \\ $$$$+\mathrm{4}{abc}^{\mathrm{4}} +\mathrm{8}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \\ $$$$=\mathrm{7}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{4}{abc}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)+\mathrm{8}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \\ $$

Commented by York12 last updated on 27/Feb/24

$$\mathrm{Perfect}\:\mathrm{sir}\:! \\ $$

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