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Question-159035




Question Number 159035 by HongKing last updated on 12/Nov/21
Commented by mr W last updated on 12/Nov/21
no unique solution possible.  we can select a very small value for  z=δ with δ→0. then we have  m=min(×, ×, ×) →0  i.e. m doesn′t exist!    i think the question should be  m=max(×,×,×)
nouniquesolutionpossible.wecanselectaverysmallvalueforz=δwithδ0.thenwehavem=min(×,×,×)0i.e.mdoesntexist!ithinkthequestionshouldbem=max(×,×,×)
Commented by HongKing last updated on 12/Nov/21
no there is solution for my problem  my dear Ser
nothereissolutionformyproblemmydearSer
Commented by mr W last updated on 12/Nov/21
then please share your solution:  how large is m?  please don′t just say, but also do!  otherwise it′s not fair from you,  because i always tell you my complete  solution, but you just say you have  the solution without telling how the  solution is.
thenpleaseshareyoursolution:howlargeism?pleasedontjustsay,butalsodo!otherwiseitsnotfairfromyou,becauseialwaystellyoumycompletesolution,butyoujustsayyouhavethesolutionwithouttellinghowthesolutionis.
Commented by HongKing last updated on 12/Nov/21
assume that → m ≥ 1  then we have (2ab^2 +b^3 c)(2bc^2 +c^3 a)(2ca^2 +a^3 b)≥1  but by → AM - GM → we have:  (2ab^2 +b^3 c)(2bc^2 +c^3 a)(2ca^2 +a^3 b)  = (abc)^2 (2a+bc)(2b+ca)(2c+ab)  ≤^(AM-GM)  (((a+b+c)/3))^6 (((2(a+b+c)(ab+bc+ca))/3))^3   ≤ ((2/3))^6 (((2∙2)/3) + (((a+b+c)^2 )/(3∙3)))^3   = ((2/3))^6 (((16)/9))^3 = ((8/9))^6 < 1  then we have → m < 1  therefore the equation → x^2  + 2x + m = 0  has root → -1 ± (√(1 - m)) •
assumethatm1thenwehave(2ab2+b3c)(2bc2+c3a)(2ca2+a3b)1butbyAMGMwehave:(2ab2+b3c)(2bc2+c3a)(2ca2+a3b)=(abc)2(2a+bc)(2b+ca)(2c+ab)AMGM(a+b+c3)6(2(a+b+c)(ab+bc+ca)3)3(23)6(223+(a+b+c)233)3=(23)6(169)3=(89)6<1thenwehavem<1thereforetheequationx2+2x+m=0hasroot1±1m
Commented by mr W last updated on 12/Nov/21
i think, as long as m doesn′t exist,   equation x^2 +2x+m=0 has no root.   this is the same as:  let m=min(f(t)) with f(t)=(1/(1+t^2 ))  solve x^2 +2x+m=0
ithink,aslongasmdoesntexist,equationx2+2x+m=0hasnoroot.thisisthesameas:letm=min(f(t))withf(t)=11+t2solvex2+2x+m=0
Commented by HongKing last updated on 13/Nov/21
My dear Ser, can you tell me more  about this fraction  (1/(1+t^2 )) ? I did not understand
MydearSer,canyoutellmemoreaboutthisfraction11+t2?Ididnotunderstand
Commented by mr W last updated on 13/Nov/21
i just want to say: the function   f(t)=(1/(1+x^2 )) is always less than or equal  to 1. but it has no minimum m, i.e.  m=min(f(t)) doesn′t exist. if m  doesn′t exist, we can also not solve  x^2 +2x+m=0, because in this  equation m is not just a variable, but  a concrete value which is defined as  the minimum value of the function  f(t)=(1/(1+t^2 )). when m doesn′t exist,  the so called roots −1±(√(1−m)) don′t  exist either.  in your question m is defined as  the minimun of a function  F(a,b,c). but this function also has  no minimum m, therefore x^2 +2x+m=0  has no roots.
ijustwanttosay:thefunctionf(t)=11+x2isalwayslessthanorequalto1.butithasnominimumm,i.e.m=min(f(t))doesntexist.ifmdoesntexist,wecanalsonotsolvex2+2x+m=0,becauseinthisequationmisnotjustavariable,butaconcretevaluewhichisdefinedastheminimumvalueofthefunctionf(t)=11+t2.whenmdoesntexist,thesocalledroots1±1mdontexisteither.inyourquestionmisdefinedastheminimunofafunctionF(a,b,c).butthisfunctionalsohasnominimumm,thereforex2+2x+m=0hasnoroots.
Commented by HongKing last updated on 13/Nov/21
m  is constant, not variable
misconstant,notvariable
Commented by mr W last updated on 13/Nov/21
m  is constant which doesn′t exist.  this is elementary, i think there is  no need for further discussion about  it.
misconstantwhichdoesntexist.thisiselementary,ithinkthereisnoneedforfurtherdiscussionaboutit.
Commented by HongKing last updated on 13/Nov/21
When we assume that m is greater,  it leads to wrong result,  then m must be less 1,  this is the main idea..
Whenweassumethatmisgreater,itleadstowrongresult,thenmmustbeless1,thisisthemainidea..
Commented by HongKing last updated on 13/Nov/21
Again..  The equation:  x^2 +2x+m=0 → with: Δ=4(1-n)  Assume that:  m>1 → then: 2ab^2 +b^3 c>1 ; 2bc^2 +c^3 a>1 ; 2ca^2 +a^3 b>1  Now using condition:  2ab^2 +b^3 c=b^2 (2(2-b-c)+bc)=b^2 (2-b)(2-c)>1  Smilarly:  2bc^2 +c^3 a=c^2 (2-a)(2-c)>1                       2ca^2 +a^3 b=a^2 (2-a)(2-b)>1  Thus:  abc(2-a)(2-b)(2-c)>1  (∗)  Using AM-GM →  0≤a(2-a)≤1 ; 0≤b(2-b)≤1 ; 0≤c(2-c)≤1  Or:  abc(2-a)(2-b)(2-c)≤1  (∗∗)  By (∗) and (∗∗) we get  m≤1  and the initial  equation has root:  x_1  ; x_2  = {-1 + (√(1 - m)) ; -1 - (√(1 - m)) }
Again..Theequation:x2+2x+m=0with:Δ=4(1n)Assumethat:m>1then:2ab2+b3c>1;2bc2+c3a>1;2ca2+a3b>1Nowusingcondition:2ab2+b3c=b2(2(2bc)+bc)=b2(2b)(2c)>1Smilarly:2bc2+c3a=c2(2a)(2c)>12ca2+a3b=a2(2a)(2b)>1Thus:abc(2a)(2b)(2c)>1()UsingAMGM0a(2a)1;0b(2b)1;0c(2c)1Or:abc(2a)(2b)(2c)1()By()and()wegetm1andtheinitialequationhasroot:x1;x2={1+1m;11m}
Commented by mr W last updated on 13/Nov/21
even when m>1 there are roots from  x^2 +2x+m=0. but even whem m<1,  but when it doesn′t exist, there is no  root from x^2 +2x+m=0. i′m only  convinced when you can find the  minimum value m concretely. the  minimum of a function is a fixed  value, not a range like m≤1.
evenwhenm>1therearerootsfromx2+2x+m=0.butevenwhemm<1,butwhenitdoesntexist,thereisnorootfromx2+2x+m=0.imonlyconvincedwhenyoucanfindtheminimumvaluemconcretely.theminimumofafunctionisafixedvalue,notarangelikem1.

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