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If we define $3v^2 = ab+bc+ca$, then doesn't that mean $3v^2$ must be positive due to the exponentiation and positive $3$?

I note that if for example $a = −1, b = −2, c = 3$ we get $3v^2 = −7$! I understand that the algebra is not tricking me but I am just not convinced that two positives multiply to form a negative.

user71207
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    Without any explanation (uwv method?) the problem is meaningless. – herb steinberg Jul 02 '21 at 03:59
  • The definition you started with does not guarantee results within the real numbers... – abiessu Jul 02 '21 at 04:00
  • Here's a much simpler example of the same difficulty. If we define $v^2=a$, then for example if $a=-1$, we get $v^2=-1$. Which has no (real) solution. There are three ways out. One, allow nonreal solutions. Two, forbid negative values of $a$. Three, accept that you have no business defining $v^2=a$. To decide which path to follow, you must know why you decided to define $v^2=a$. We don't know why you decided to define $3v^2=ab+bc+ca$, so we can't help you. – Gerry Myerson Jul 02 '21 at 04:16
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    The uvw method is applied for non-negative reals. We use AM-GM (and other inequalities) to establish the results (which is why we need the terms to be non-negative). Sp, $ 3v^2 =ab+bc+ca$ cannot be negative by definition. – Calvin Lin Jul 02 '21 at 04:40

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