Re: missing the point of Proof...
- From: "Randy Poe" <poespam-trap@xxxxxxxxx>
- Date: 8 Jan 2006 18:36:40 -0800
Steve Mazza wrote:
> Hi, all. I think I am missing the point of what it means to *prove* an
> inequality.
I may not have addressed your main point, which was
"what am I supposed to be proving"? You are being
asked to show that the inequality holds whenever the
assumptions (x>0, y>0) are true. For instance, suppose
I pick x=1, y=2. Then (1/x + 1/y) = 1.5, (x+y) = 3, and
their product is 4.5 which is clearly >= 4. Is it possible
to pick x and y such that the product is less than 4?
You are showing that the answer is no: so long as you
start with positive x and positive y, the product of these
two terms will be at least 4.
As another example, there is a well-known theorem that
the arithmetic mean of two positive numbers, (a+b)/2, is at
least as large as the geometric mean, sqrt(ab).
That is, (a+b)/2 >= sqrt(ab).
(Actually, the theorem applies to any number of numbers,
not just two).
To prove this means to prove that (a+b)/2 - sqrt(ab) is
always a positive number, for any choice of a and b > 0.
The proof is simple:
(a-b)^2 >= 0 for any a, b.
a^2 - 2ab + b^2 >= 0 (multiply out)
a^2 + 2ab + b^2 >= 4ab (add 4ab to both sides)
(a+b)^2 >= 4ab
For positive a, b, both sides of this are positive. We can
take the positive square root and not change the direction
of the inequality.
(a+b) >= 2 sqrt(ab)
And that completes the proof. Note that if either a or b
were negative we could not do this last step.
- Randy
.
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