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watchmath
 4 years ago
Given that \(a+b+c=0\). Find
\[\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\]
watchmath
 4 years ago
Given that \(a+b+c=0\). Find \[\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this problem is a bit lengthy to type out....but the process would be to find a common denom.....so that you can add the numerators. lots of algebra. you will have some cancellations in the numerator once you combine like terms.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a = 1,b=2,c=3 (52+1/3)(1/5 1/2 +3) = (10/3)(27/10) = 9 a = 1,b=3,c=4 (75/3+1/2)(1/7 3/5 +2) = (35/6)(54/35) = 9 this is only 2 cases but Im guessing it will always equal 9

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im doing this in my head right now otherwise i'd try to do all the algebra to prove it

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.1still remember me dumbcow? :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Very cool problem that has a very interesting proof that isn't just computationally based. First, the AMGM inequality has a very natural extension, the AMGMHM inequality, where HM is the harmonic mean. In three variables, this is \[\frac{x+y+z}{3}\ge (xyz)^\frac13 \ge \frac3{\frac1x+\frac1y+\frac1z}\] Setting \[x=\frac{bc}{a},y=\frac{ca}{b},z=\frac{ab}{c}\] Your equation becomes just \[(x+y+z)\left(\frac1x+\frac1y+\frac1z\right)\] From AMGMHM, you have \[\frac{x+y+z}{3}\ge \frac3{\frac1x+\frac1y+\frac1z} \implies (x+y+z)\left(\frac1x+\frac1y+\frac1z\right) \ge 9\] Another property of this inequality is that equality is only achieved when x=y=z. Setting \[\frac{bc}{a}=\frac{ca}{b}=\frac{ab}{c}=k\] Solving each of the three equalities gives a system \[akb+c=a+bkc=a+b+ck=0\] Summing the three gives \[ak+bk+ck=k(a+b+c)=0\] Since k cannot equal zero (since then a=b=c=0, but you can't divide by zero), then a+b+c=0. All of these final few steps are reversible, which gives the answer that the equation is always equal to 9.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Actually, there is a flaw in this proof. Give me a little bit and I'll fix it.

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.1for AMGm, x,y,z need to be non negative
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