Question

Given that \(a+b+c=0\). Find
\[\left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\]

Answer 1

this 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.

Answer 2

a = 1,b=2,c=-3
(5-2+1/3)(1/5 -1/2 +3) = (10/3)(27/10) = 9

a = 1,b=3,c=-4
(7-5/3+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

Answer 3

im doing this in my head right now otherwise i'd try to do all the algebra to prove it

Answer 4

still remember me dumbcow? :D

Answer 5

haha yeah kinda

Answer 6

Very cool problem that has a very interesting proof that isn't just computationally based. First, the AM-GM inequality has a very natural extension, the AM-GM-HM 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 \[x=\frac{b-c}{a},y=\frac{c-a}{b},z=\frac{a-b}{c}\] Your equation becomes just \[(x+y+z)\left(\frac1x+\frac1y+\frac1z\right)\] From AM-GM-HM, 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 \[\frac{b-c}{a}=\frac{c-a}{b}=\frac{a-b}{c}=k\] Solving each of the three equalities gives a system \[ak-b+c=a+bk-c=-a+b+ck=0\] 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.

Answer 7

Actually, there is a flaw in this proof. Give me a little bit and I'll fix it.

Answer 8

for AM-Gm, x,y,z need to be non negative