Polytechnic University of Bucharest
Question
Asked 21st Jul, 2013
Does anybody have or know where to find a proof for the following inequality in generalized geometric means?
Suppose you have n numbers on the unit interval [0,1]. Also suppose that you have n weights on the unit interval such that the weights add up to 1. Now build the generalized geometric mean of the n numbers x given the weights w, i.e. the product of all x^w. Also build the geometric mean of all duals of x given the same weights, i.e. the product of all (1-x)^w. Is it true that the sum of both geometric means is equal or less than 1? For all special cases and a large number of random cases it turned out to be true, so this is my conjecture. Probably not too difficult to prove, but I need some hints where to look or how it relates to other well-known theorems.
Most recent answer
Dear Michael,
The proof of Jensen inequality for convex functions (the discrete case) can be handled by induction and is very simple too. It solves not only our problem, but also the integral form of Jensen inequality and many other basic results (such as Holder inequality, etc.). See the book of Walter Rudin "Real and complex analysis"...
Regards,
Octav
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Popular answers (1)
ISSAC Corporation
One proof seems like a straightforward use of the Arithmetic-Geometric Inequality, that AM(w,s)>=GM(w,s) for the vector s>0 and 0<w<1, sum(w)=1. I believe that your problem additionally has s<1, but it doesn't matter. State the AG Inequalities for both s and 1-s, then add the respective left- and right-hand sides of each together. I think that yields your result. Sorry that my proficiency with algebraic expression typing is so shabby, but I hope this helps.
The geometric programming literature might be helpful here. Duffin, Peterson and Zener's book (1977) has foundational theorems and references to Holder's Inequality which your proof could leverage.
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All Answers (15)
Technische Universität Berlin
Many of the inequalities between sums of products and products of sums (like the inequality between geometrical and arithmetical mean, Hölder inequality etc.) can be proved by means of the convexity or concavity of some function f(x_1,x_2,...x_n). In your case you want to prove the inequality $\prod x_i^{w_i} + \prod (1-x_i)^{w_i}\leq 1$. For $f(x_1,...,x_n)=x_1^{w_1} \cdot ... \cdot x_n^{w_n}$ this inequality can be written:
$f(s_1, ... s_n)+ f(1-s_1, ..., 1-s_n)\leq 1 $, and this is equivalent to
$1/2 (f(s_1, ... s_n)+ f(1-s_1, ..., 1-s_n)) \leq 1/2 =f(1/2,...,1/2) $. The inequality would follow, if you can prove the concavity of $f$. -- In the book 'Inequalities' by Hardy and Littlewood you will find this method and other methods of proof for such inequality. I do not know whether your specific example occurs in the book.
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ISSAC Corporation
One proof seems like a straightforward use of the Arithmetic-Geometric Inequality, that AM(w,s)>=GM(w,s) for the vector s>0 and 0<w<1, sum(w)=1. I believe that your problem additionally has s<1, but it doesn't matter. State the AG Inequalities for both s and 1-s, then add the respective left- and right-hand sides of each together. I think that yields your result. Sorry that my proficiency with algebraic expression typing is so shabby, but I hope this helps.
The geometric programming literature might be helpful here. Duffin, Peterson and Zener's book (1977) has foundational theorems and references to Holder's Inequality which your proof could leverage.
4 Recommendations
Edumetrics Research & Development, Germany
Thanks a lot to all respondents. Yes, the Arithmetic-Geometric Inequality (with weights) was the right hint. It can be simply proved by Jensen's famous Inequality (for the discrete case), see Wikipedia. Of course,a direct proof would have to dig deeper (thanks to Alexander Kukush), but for my purpose a reference to Jensen will suffice.
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Polytechnic University of Bucharest
Octav Olteanu, University Politehnica of Bucharest
The inequality is true, thanks to Jensen inequality applied to each generalized geometric means. The condition Sum w=1 is essential. Otherwise, for arbitrary w in (0,1), the inequality is not true.
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Institut de Mécanique des Fluides de Toulouse
Hello Paul Hubert VOSSEN: R
Your question seems related to a theroem known as Jensen's inequality for convex functions. I have used it in a probability course (and in my research on stochastic conduction equations) to demonstrate that the arithmetic mean < geom mean < harmon mean. The more general case of power averages is also treated in my 2 page note (I call it Holder average, with an umlaut on the "o" of Holder). The attached note is short (2 pages): I hope you can decipher it (it is in french). Best regards, Rachid ABABOU, Pr. <ababou@imft.fr>
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Institut de Mécanique des Fluides de Toulouse
Hello Paul Hubert VOSSEN:
Your question seems related to a theroem known as Jensen's inequality for convex functions. I have used it in a probability course (and in my research on stochastic conduction equations) to demonstrate that the arithmetic mean < geom mean < harmon mean. The more general case of power averages is also treated in my 2 page note (I call it Holder average, with an umlaut on the "o" of Holder). The attached note is short (2 pages): I hope you can decipher it (it is in french). Best regards, Rachid ABABOU, Pr. <ababou@imft.fr>
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Polytechnic University of Bucharest
Octav Olteanu, University Politehnica of Bucharest
Using the case when in Jensen (discrete) inequality we have equality, one can prove easily that in the present inequality equality occurs if and only if all the x numbers are equal to a common value.
Cornell University
A direct proof doesn't need any deep arguments, just a simple argument
by induction on n. The result is true for n=2 for simple reasons; suppose it true for n >= 1,
and consider a case with n+1 x's and w's, indexed 0 through n. If some w_i
is zero, this reduces to the n-dimensional case. If not, W' = 1 - w_0 is
positive, and x_0^w_0 ... x_n^w_n = x_0^w_0 X'^W', where
X' = x_1^w'_1 ... x_n^w'_n, with w'_j = w_j/W'..
Similarly, (1-x_0)^w_0 ... (1-x_n)^w_n = (1-x_0)^w_0 Y'^W' <= (1-x_0)^w_0 (1-X')^W'
using the induction hypothesis on X' and Y'. Now the result follows from the n=2 case.
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University of Auckland
One can assume that all the w values are rational with a common denominator
because if the result were false, possibly with irrational w values, it would also be false if the w's were replaced by close rational approximations. Given that the w's are rational m/n where n is constant then x^(m/n) is the product of m equal copies of
x^(1/n). Hence the result follows from the classical geometrical mean of n x values
+ the classical geometric mean of corresponding (1-x) values. Each of the two geometrical means is less than or equal to the corresponding arithmetic means where the sum of the two terms is exactly 1.
voestalpine Stahl Gmbh and Johannes Kepler University
Hallo, a very good book on inequalities is "The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities ( Problem Books Series... von J. Michael Steele von Cambridge University Press", perhaps you can find some hints how to solve your problem. It is also quite inexpensive and readable!
Best regards, Stefan
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Google Inc.
The simple argument is that the geometric mean of a series of positive numbers is always less than or equal to their arithmetic mean. Then:
Product ( x_i^(w_i) ) <= 1 / n Sum (x_i w_i)
Product ( (1 - x_i)^(w_i)) <= 1 / n Sum ( (1 - x_i) w_i) = 1 - Sum(x_i w_i) / n
If you sum up the sides, the average terms cancel out and you are left with your conjecture. So yes, it is true.
University of Cambridge
Another way to prove your "conjecture" is using the general version of the Young's inequality; which says if \Sum {1/p_i} = 1 , p_i>1, 1<=i<=n, then
\Product {a_i} <= \Sum {a_i^{p_i}/p_i}.
To prove the desired result just apply to \Product { x_i^(w_i) } + \Product { (1 - x_i)^(w_i)} with p_i = 1/w_i and a_i = x_i^{w_i} , (1-x_i)^{w_i}. One has
\Product { x_i^(w_i) } + \Product {(1 - x_i)^(w_i)} <= \Sum {x_i/w_i} + \Sum {(1-x_i)/w_i} = \Sum {w_i} = 1.
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Polytechnic University of Bucharest
Dear Michael,
The proof of Jensen inequality for convex functions (the discrete case) can be handled by induction and is very simple too. It solves not only our problem, but also the integral form of Jensen inequality and many other basic results (such as Holder inequality, etc.). See the book of Walter Rudin "Real and complex analysis"...
Regards,
Octav
1 Recommendation
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