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Prove this is greater than or equal to 8 
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#1
Sep2411, 01:12 AM

P: 24

1. The problem statement, all variables and given/known data
Show that, if a, b, c >0 and a+b+c = 1, then, ([itex]\frac{1}{a}[/itex]1)([itex]\frac{1}{b}[/itex]1)([itex]\frac{1}{c}[/itex]1) [itex]\geq[/itex] 8 2. Relevant equations The usual math is ok 3. The attempt at a solution I multiplied them out to get: ([itex]\frac{1}{abc}\frac{1}{ac}\frac{1}{bc}+\frac{1}{c}\frac{1}{ab}+\frac{1}{a}+\frac{1}{b}[/itex]1) = ([itex]\frac{1}{abc}\frac{b}{abc}\frac{a}{abc}+\frac{ab}{abc}\frac{c}{abc}+\frac{bc}{abc}+\frac{ac}{abc}[/itex]1) = ([itex]\frac{1ba+abc+bc+acabc}{abc}[/itex]) = ([itex]\frac{ab+bc+ac}{abc}[/itex]) = ([itex]\frac{1}{a}+\frac{1}{b}+\frac{1}{c}[/itex]), .........as 1(a+b+c) = 0 abc is a small number as a,b,c are all less than 1, and it is and order smaller than anything in the numerator. OR, since a, b, c < 1, ([itex]\frac{1}{a}+\frac{1}{b}+\frac{1}{c}[/itex]) > 3 Now I don't know what to do. I'm not even sure I'm on the right track... There may be some other logic I should be using to make this easier, but I don't see it. 


#2
Sep2411, 01:47 PM

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#3
Sep2411, 03:31 PM

P: 24

so, use c = 1  a  b, substitute that into the first equation, and see what happens? 


#4
Sep2411, 06:34 PM

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Prove this is greater than or equal to 8
That's what I had in mind, yes.
I tried that, and was able to factor things a lot, but I wasn't able to show that the expression on the left is >= 8. Where you might go next is to consider w = f(x, y) = (1/x  1)(1/y  1)( (x + y)/(1 x  y)). Use calculus to show that there is a global minimum, and that the minimum value is 8. 


#5
Sep2411, 07:21 PM

P: 24

[itex](\frac{1}{a}1)[/itex] [itex]\geq[/itex] 2, as, if this is done, without loss of generality, it would apply to each, and 2x2x2 [itex]\geq[/itex] 8 


#6
Sep2411, 09:06 PM

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#7
Sep2411, 09:37 PM

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P: 1,662

I can't judge from your title or first post what level of analysis you have had already. So I'll propose a couple of things:
 the straightforward one if you've had multivariate calculus is to use Lagrange multipliers, since you have a function and a constraint equation in three variables; you will be able to show easily that the critical point occurs where a = b = c ; you would then reduce the function to one with two variables and use the Second Partial Derivative Test to show that the value 8 is a minimum [the Test with three variable is messier]; this is, I believe, essentially what Mark44 is suggesting  a noncalculus approach would be to consider that a + b + c = 1 in the first octant is an equilateral triangular region, the portion of the plane bounded by the coordinate planes; the product in question is plainly infinite at the edges of the triangle, so the interior must contain a minimum for the product; the threefold symmetry of the region suggests that the "special point" lies where a = b = c = 1/3 ; that gives the "right number" for the product, but what is the behavior of the function?; fix one variable and shift one of the others by a small amount, which automatically shifts the remaining variable; show that any deviation from the special point must give a larger product 


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