Prove this is greater than or equal to 8


by Chinnu
Tags: analysis, proof
Chinnu
Chinnu is offline
#1
Sep24-11, 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{1-b-a+ab-c+bc+ac-abc}{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.
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Mark44
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#2
Sep24-11, 01:47 PM
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P: 20,999
Quote Quote by Chinnu View Post
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{1-b-a+ab-c+bc+ac-abc}{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,
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 Quote by Chinnu View Post


([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.
Chinnu
Chinnu is offline
#3
Sep24-11, 03:31 PM
P: 24
Quote Quote by Mark44 View Post
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?

Mark44
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#4
Sep24-11, 06:34 PM
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P: 20,999

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.
Chinnu
Chinnu is offline
#5
Sep24-11, 07:21 PM
P: 24
Quote Quote by Mark44 View Post
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:

[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
Mark44
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#6
Sep24-11, 09:06 PM
Mentor
P: 20,999
Quote Quote by Chinnu View Post
I will try this. Also, would it be possible to see if:

[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
The inequality above is true for 0 < x <= 1/3.
dynamicsolo
dynamicsolo is offline
#7
Sep24-11, 09:37 PM
HW Helper
P: 1,664
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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