Prove this is greater than or equal to 8

by Chinnu
Tags: analysis, proof
 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, ($\frac{1}{a}$-1)($\frac{1}{b}$-1)($\frac{1}{c}$-1) $\geq$ 8 2. Relevant equations The usual math is ok 3. The attempt at a solution I multiplied them out to get: ($\frac{1}{abc}-\frac{1}{ac}-\frac{1}{bc}+\frac{1}{c}-\frac{1}{ab}+\frac{1}{a}+\frac{1}{b}$-1) = ($\frac{1}{abc}-\frac{b}{abc}-\frac{a}{abc}+\frac{ab}{abc}-\frac{c}{abc}+\frac{bc}{abc}+\frac{ac}{abc}$-1) = ($\frac{1-b-a+ab-c+bc+ac-abc}{abc}$) = ($\frac{ab+bc+ac}{abc}$) = ($\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$), .........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, ($\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$) > 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.
Mentor
P: 21,311
 Quote by Chinnu 1. The problem statement, all variables and given/known data Show that, if a, b, c >0 and a+b+c = 1, then, ($\frac{1}{a}$-1)($\frac{1}{b}$-1)($\frac{1}{c}$-1) $\geq$ 8 2. Relevant equations The usual math is ok 3. The attempt at a solution I multiplied them out to get: ($\frac{1}{abc}-\frac{1}{ac}-\frac{1}{bc}+\frac{1}{c}-\frac{1}{ab}+\frac{1}{a}+\frac{1}{b}$-1) = ($\frac{1}{abc}-\frac{b}{abc}-\frac{a}{abc}+\frac{ab}{abc}-\frac{c}{abc}+\frac{bc}{abc}+\frac{ac}{abc}$-1) = ($\frac{1-b-a+ab-c+bc+ac-abc}{abc}$) = ($\frac{ab+bc+ac}{abc}$) = ($\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$), .........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,
According to what you wrote, a + b + c = 1. You haven't used this fact. If you do, you can get an expression that doesn't involve c.
 Quote by Chinnu ($\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$) > 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.
P: 24
 Quote by Mark44 According to what you wrote, a + b + c = 1. You haven't used this fact. If you do, you can get an expression that doesn't involve c.

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

 Mentor P: 21,311 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.
P: 24
 Quote by Mark44 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.
I will try this. Also, would it be possible to see if:

$(\frac{1}{a}-1)$ $\geq$ 2,

as, if this is done, without loss of generality, it would apply to each, and 2x2x2 $\geq$ 8
Mentor
P: 21,311
 Quote by Chinnu I will try this. Also, would it be possible to see if: $(\frac{1}{a}-1)$ $\geq$ 2, as, if this is done, without loss of generality, it would apply to each, and 2x2x2 $\geq$ 8
The inequality above is true for 0 < x <= 1/3.
 HW Helper 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 non-calculus 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 three-fold 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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