Prove a = b = c
Date: 01/27/2002 at 15:05:55 From: Jim Chang Subject: Proof When a^2 + b^2 + c^2 = ab + bc + ca and abc does not equal 0, prove that a = b = c. I tried aa + bb + cc = ab + bc + ca and then substituting all a's for b's: bb + bb + cc = bb + bc + cb, giving me bb + cc = 2bc, which is correct, but I need algebraic proof. Thanks!
Date: 01/28/2002 at 03:07:42 From: Doctor Floor Subject: Re: Proof Hi, Jim, Thanks for your question. From abc unequal to zero, we see that none of a,b,c is equal to zero. Now let's suppose that a <= b <= c. Then we can find nonnegative integers t and u satisfying b = a + t and c = a + u. Substitution yields a^2 + b^2 + c^2 = 3a^2 + 2(t+u)a + t^2 + u^2 ab + bc + ba = 3a^2 + 2(t+u)a + tu So we derive from a^2+b^2+c^2 = ab+bc+ca that t^2 + u^2 = tu [*] We now must show that t = u = 0. If t = 0, and we substitute in [*], we find that u^2 = 0, and thus u = 0. So it is enough to show that one of the two must be 0. Suppose that both t and u are positive. Then it is not difficult to see: * if t > u then t^2 > ut and thus also t^2 + u^2 > ut, * if t < u then t^2 + u^2 > ut in the same way. * if t = u, clearly t^2 + u^2 = 2ut > ut. So that leads to a contradiction. And thus t = u = 0, and also a = b = c. If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
Date: 02/04/2002 at 18:42:57 From: Doctor Allan Subject: Re: Proof Hello Jim, I would like to offer another solution to your problem besides the good solution that Dr. Floor already provided to you. If we try to multiply the equation by 2 we get 2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca Rewrite this as 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca = 0 a^2 + a^2 + b^2 + b^2 + c^2 + c^2 - 2ab - 2bc - 2ca = 0 Rearranging the terms yields a^2 + b^2 - 2ab + b^2 + c^2 - 2bc + c^2 + a^2 - 2ca = 0 (a-b)^2 + (b-c)^2 + (c-a)^2 = 0 When is it the case that this is zero? - Doctor Allan, The Math Forum http://mathforum.org/dr.math/
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