| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
As part of a more general problem I needed to prove the following:
(xi are positive reals, i = 1 to n)
if Product(xi) is fixed
then Sum(xi) attains its least value when the xi are equal.
Can anyone supply a proof? I'm sure I shall blush when someone provides a
simple one.
For n=2 it follows immediately from:
(x1 + x2)^2 = 4*x1*x2 + (x1 - x2)^2
Dick
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1263.1 | TRACE::GILBERT | Ownership Obligates | Mon Jul 02 1990 15:18 | 10 | |
Suppose there are two unequal xi. Without loss of generality, call them x1 and x2. replacing them with x1' = x2' = sqrt( x1 x2 ) leaves the product unchanged, but reduces the sum of the xi, viz: 0 < (x1 - x2)^2 (since they're unequal) 0 < x1^2 - 2 x1 x2 + x2^2 4 x1 x2 < x1^2 + 2 x1 x2 + x2^2 4 x1 x2 < (x1 + x2)^2 2 * sqrt( x1 x2 ) < x1 + x2 x1' + x2' < x1 + x2 | |||||
| 1263.2 | ALLVAX::JROTH | It's a bush recording... | Mon Jul 02 1990 16:31 | 18 | |
Another standard way for this class of problem is via Lagrange
multipliers.
To minimize S = SUM(x[i]) subject to PROD(x[i]) = C, minimize instead
S + L*(PROD(x[i])-C), over the x's and L, the Lagrange multiplier.
Taking partial derivatives with respect to the x's gives
1 + L*(PROD(x[i]))/x[j] = 0, j = 1,2,...
x[j] = -L*C, j = 1, 2, ...
Which implies that the x's are equal at the constrained extremum.
Still, I like the proof in .1 much better.
- Jim
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| 1263.3 | Paul Erdos (sp?) would approve! | IOSG::CARLIN | Dick Carlin IOSG, Reading, England | Tue Jul 03 1990 11:11 | 2 |
Many thanks, Peter. A very neat proof.
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