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k a My Retirement & New Leadership at AoPS
rrusczyk   1571
N Mar 26, 2025 by SmartGroot
I write today to announce my retirement as CEO from Art of Problem Solving. When I founded AoPS 22 years ago, I never imagined that we would reach so many students and families, or that we would find so many channels through which we discover, inspire, and train the great problem solvers of the next generation. I am very proud of all we have accomplished and I’m thankful for the many supporters who provided inspiration and encouragement along the way. I'm particularly grateful to all of the wonderful members of the AoPS Community!

I’m delighted to introduce our new leaders - Ben Kornell and Andrew Sutherland. Ben has extensive experience in education and edtech prior to joining AoPS as my successor as CEO, including starting like I did as a classroom teacher. He has a deep understanding of the value of our work because he’s an AoPS parent! Meanwhile, Andrew and I have common roots as founders of education companies; he launched Quizlet at age 15! His journey from founder to MIT to technology and product leader as our Chief Product Officer traces a pathway many of our students will follow in the years to come.

Thank you again for your support for Art of Problem Solving and we look forward to working with millions more wonderful problem solvers in the years to come.

And special thanks to all of the amazing AoPS team members who have helped build AoPS. We’ve come a long way from here:IMAGE
1571 replies
rrusczyk
Mar 24, 2025
SmartGroot
Mar 26, 2025
k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
nice problem
hanzo.ei   2
N 2 minutes ago by Lil_flip38
Source: I forgot
Let triangle $ABC$ be inscribed in the circumcircle $(O)$ and circumscribed about the incircle $(I)$, with $AB < AC$. The incircle $(I)$ touches the sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. A line through $I$, perpendicular to $AI$, intersects $BC$, $CA$, and $AB$ at $X$, $Y$, and $Z$, respectively. The line $AI$ meets $(O)$ at $M$ (distinct from $A$). The circumcircle of triangle $AYZ$ intersects $(O)$ at $N$ (distinct from $A$). Let $P$ be the midpoint of the arc $BAC$ of $(O)$. The line $AI$ cuts segments $DF$ and $DE$ at $K$ and $L$, respectively, and the tangents to the circle $(DKL)$ at $K$ and $L$ intersect at $T$. Prove that $AT \perp BC$.
2 replies
hanzo.ei
Yesterday at 5:58 PM
Lil_flip38
2 minutes ago
Fixed point config on external similar isosceles triangles
Assassino9931   2
N 5 minutes ago by bin_sherlo
Source: Bulgaria Spring Mathematical Competition 2025 10.2
Let $AB$ be an acute scalene triangle. A point \( D \) varies on its side \( BC \). The points \( P \) and \( Q \) are the midpoints of the arcs \( \widehat{AB} \) and \( \widehat{AC} \) (not containing \( D \)) of the circumcircles of triangles \( ABD \) and \( ACD \), respectively. Prove that the circumcircle of triangle \( PQD \) passes through a fixed point, independent of the choice of \( D \) on \( BC \).
2 replies
Assassino9931
Today at 12:41 PM
bin_sherlo
5 minutes ago
VERY HARD MATH PROBLEM!
slimshadyyy.3.60   17
N 12 minutes ago by GreekIdiot
Let a ≥b ≥c ≥0 be real numbers such that a^2 +b^2 +c^2 +abc = 4. Prove that
a+b+c+(√a−√c)^2 ≥3.
17 replies
slimshadyyy.3.60
Yesterday at 10:49 PM
GreekIdiot
12 minutes ago
Question 2
Valentin Vornicu   87
N 30 minutes ago by ItsBesi
Consider five points $ A$, $ B$, $ C$, $ D$ and $ E$ such that $ ABCD$ is a parallelogram and $ BCED$ is a cyclic quadrilateral. Let $ \ell$ be a line passing through $ A$. Suppose that $ \ell$ intersects the interior of the segment $ DC$ at $ F$ and intersects line $ BC$ at $ G$. Suppose also that $ EF = EG = EC$. Prove that $ \ell$ is the bisector of angle $ DAB$.

Author: Charles Leytem, Luxembourg
87 replies
Valentin Vornicu
Jul 25, 2007
ItsBesi
30 minutes ago
No more topics!
The Karamata inequality
darij grinberg   41
N Sep 7, 2014 by Fu_Manchu
Source: Majorization Theory
Since I was asked about it I'm posting it.

We begin with a definition:

1. Let $ x_1$, $ x_2$, ..., $ x_n$, $ y_1$, $ y_2$, ..., $ y_n$ be arbitrary real numbers satisfying $ x_1 \geq x_2 \geq ... \geq x_n$ and $ y_1 \geq y_2 \geq ... \geq y_n$. Now, if we have all the following conditions fulfilled:

$ x_1 \geq y_1$;
$ x_1 + x_2 \geq y_1 + y_2$;
$ x_1 + x_2 + x_3 \geq y_1 + y_2 + y_3$;
...
generally $ x_1 + x_2 + ... + x_k \geq y_1 + y_2 + ... + y_k$ for any natural k with $ 1 \leq k \leq n - 1$;
and $ x_1 + x_2 + ... + x_n = y_1 + y_2 + ... + y_n$

(mind the equality sign in the last condition; it is not a $ \geq$ sign!), then we say that the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$. We write this in the form $ \left( x_1;\;x_2;\;...;\;x_n\right) \succ \left( y_1;\;y_2;\;...;\;y_n\right)$.

Now, if instead of

$ x_1 \geq y_1$;
$ x_1 + x_2 \geq y_1 + y_2$;
$ x_1 + x_2 + x_3 \geq y_1 + y_2 + y_3$;
...
generally $ x_1 + x_2 + ... + x_k \geq y_1 + y_2 + ... + y_k$ for any natural k with $ 1 \leq k \leq n - 1$;
and $ x_1 + x_2 + ... + x_n = y_1 + y_2 + ... + y_n$,

we have the conditions

$ x_1 \leq y_1$;
$ x_1 + x_2 \leq y_1 + y_2$;
$ x_1 + x_2 + x_3 \leq y_1 + y_2 + y_3$;
...
generally $ x_1 + x_2 + ... + x_k \leq y_1 + y_2 + ... + y_k$ for any natural k with $ 1 \leq k \leq n - 1$;
and $ x_1 + x_2 + ... + x_n = y_1 + y_2 + ... + y_n$

fulfilled (i. e., the same conditions with all $ \geq$'s replaced by $ \leq$'s), then we say that the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ minorizes the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$. We write this in the form $ \left( x_1;\;x_2;\;...;\;x_n\right) \prec \left( y_1;\;y_2;\;...;\;y_n\right)$. Of course, this is equivalent to $ \left( y_1;\;y_2;\;...;\;y_n\right) \succ \left( x_1;\;x_2;\;...;\;x_n\right)$.

2. Thus we have defined the notions "majorizes" and "minorizes" only for non-increasing arrays (i. e. for arrays satisfying $ x_1 \geq x_2 \geq ... \geq x_n$ and $ y_1 \geq y_2 \geq ... \geq y_n$). Now, assume that $ x_1$, $ x_2$, ..., $ x_n$, $ y_1$, $ y_2$, ..., $ y_n$ are just arbitrary real numbers, without any conditions. Then, let $ \left(X_1;\;X_2;\;...;\;X_n\right)$ be the non-increasing permutation of the array $ \left(x_1;\;x_2;\;...;\;x_n\right)$, i. e. the permutation of the array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ that satisfies $ X_1 \geq X_2 \geq ... \geq X_n$. Similarly, let $ \left(Y_1;\;Y_2;\;...;\;Y_n\right)$ be the non-increasing permutation of the array $ \left(y_1;\;y_2;\;...;\;y_n\right)$, i. e. the permutation of the array $ \left(y_1;\;y_2;\;...;\;y_n\right)$ that satisfies $ Y_1 \geq Y_2 \geq ... \geq Y_n$. The number arrays $ \left(X_1;\;X_2;\;...;\;X_n\right)$ and $ \left(Y_1;\;Y_2;\;...;\;Y_n\right)$ are both non-increasing, and hence we have defined majorization for such arrays.

Then, we say that the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$ if and only if the number array $ \left(X_1;\;X_2;\;...;\;X_n\right)$ majorizes the number array $ \left(Y_1;\;Y_2;\;...;\;Y_n\right)$. In this case, we write $ \left( x_1;\;x_2;\;...;\;x_n\right) \succ \left( y_1;\;y_2;\;...;\;y_n\right)$.

Similarly, we say that the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ minorizes the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$ if and only if the number array $ \left(X_1;\;X_2;\;...;\;X_n\right)$ minorizes the number array $ \left(Y_1;\;Y_2;\;...;\;Y_n\right)$. In this case, we write $ \left( x_1;\;x_2;\;...;\;x_n\right) \prec \left( y_1;\;y_2;\;...;\;y_n\right)$. Again this is equivalent to $ \left( y_1;\;y_2;\;...;\;y_n\right) \succ \left( x_1;\;x_2;\;...;\;x_n\right)$.

3. So we have defined the terms "majorize" and "minorize" for any two number arrays. It should be noted that majorization is a partial order on the set of number arrays, not a total order - i. e., not for every pair of two number arrays $ \left(x_1;\;x_2;\;...;\;x_n\right)$ and $ \left(y_1;\;y_2;\;...;\;y_n\right)$ one can say that either the first one majorizes the second one, or the second one majorizes the first one. It often happens that none of the arrays majorizes or minorizes the other one. But sometimes when you have some special arrays, you can prove that one of them majorizes the other one.

4. Now, the Karamata inequality, also called the Majorization Inequality or the Hardy-Littlewood inequality, states that if $ x_1$, $ x_2$, ..., $ x_n$, $ y_1$, $ y_2$, ..., $ y_n$ are $ 2n$ reals from an interval $ I\subseteq\mathbb{R}$ such that the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$, and $ f: I\to\mathbb{R}$ is any convex function, then

$ f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right) \geq f\left( y_1\right) + f\left( y_2\right) + ... + f\left( y_n\right)$.

If $ f: I\to\mathbb{R}$ is a concave function instead, then we instead have

$ f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right) \leq f\left( y_1\right) + f\left( y_2\right) + ... + f\left( y_n\right)$.

If the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ minorizes the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$ instead of majorizing it, then both inequalities are reversed.

5. The Jensen inequality for real numbers is a special case of the Karamata inequality. In fact, if $ m = \frac {x_1 + x_2 + ... + x_n}{n}$ is the arithmetic mean of the numbers $ x_1$, $ x_2$, ..., $ x_n$, then it is easy to show that the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes the number array $ \left(m;\;m;\;...;\;m\right)$. Hence, the Karamata inequality yields:

If f(x) is any convex function, then

$ f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right) \geq n f\left( m\right)$,

i. e.

$ \frac {f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right)}{n} \geq f\left( m\right)$.

If f(x) is a concave function instead, then we instead have

$ f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right) \leq n f\left( m\right)$,

i. e.

$ \frac {f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right)}{n} \leq f\left( m\right)$.

Of course, this is exactly the Jensen inequality for $ n$ reals.

6. The definition of majorizing and minorizing number arrays given above is somewhat unsatisfying from an intuitive point of view, since it does not help one to imagine how an array majorizing another array looks like. Unfortunately, this is partly inherent to the notion of majorization, which indeed is quite unintuitive. For a - rather facile - visualization of the notion, you can imagine that a number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes a number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$ if the two arrays have the same sum of numbers, but the numbers of the first array are set wider apart than those of the second array, while those of the second array lie closer together. From this intuitive viewpoint, it is clear why the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes the number array $ \left(m;\;m;\;...;\;m\right)$, where $ m = \frac {x_1 + x_2 + ... + x_n}{n}$ is the arithmetic mean of the numbers $ x_1$, $ x_2$, ..., $ x_n$: In fact, the two number arrays have the same sum of elements, but the elements of the second number array lie nearer to each other (they are all equal). Alas, this viewpoint does not help one to really understand what majorization is about.

Well, I know there is more to say. For instance, the Karamata inequality has a kind of converse, but I am not sure how it is formulated, so I leave this to the other MathLinkers more used to inequalities.

Darij
41 replies
darij grinberg
Aug 5, 2004
Fu_Manchu
Sep 7, 2014
The Karamata inequality
G H J
Source: Majorization Theory
The post below has been deleted. Click to close.
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darij grinberg
6555 posts
#1 • 32 Y
Y by jatin, az360, hctb00, dan23, dantx5, cobbler, dizzy, Einstein314, champion999, 62861, Kgxtixigct, richrow12, Adventure10, Mango247, and 18 other users
Since I was asked about it I'm posting it.

We begin with a definition:

1. Let $ x_1$, $ x_2$, ..., $ x_n$, $ y_1$, $ y_2$, ..., $ y_n$ be arbitrary real numbers satisfying $ x_1 \geq x_2 \geq ... \geq x_n$ and $ y_1 \geq y_2 \geq ... \geq y_n$. Now, if we have all the following conditions fulfilled:

$ x_1 \geq y_1$;
$ x_1 + x_2 \geq y_1 + y_2$;
$ x_1 + x_2 + x_3 \geq y_1 + y_2 + y_3$;
...
generally $ x_1 + x_2 + ... + x_k \geq y_1 + y_2 + ... + y_k$ for any natural k with $ 1 \leq k \leq n - 1$;
and $ x_1 + x_2 + ... + x_n = y_1 + y_2 + ... + y_n$

(mind the equality sign in the last condition; it is not a $ \geq$ sign!), then we say that the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$. We write this in the form $ \left( x_1;\;x_2;\;...;\;x_n\right) \succ \left( y_1;\;y_2;\;...;\;y_n\right)$.

Now, if instead of

$ x_1 \geq y_1$;
$ x_1 + x_2 \geq y_1 + y_2$;
$ x_1 + x_2 + x_3 \geq y_1 + y_2 + y_3$;
...
generally $ x_1 + x_2 + ... + x_k \geq y_1 + y_2 + ... + y_k$ for any natural k with $ 1 \leq k \leq n - 1$;
and $ x_1 + x_2 + ... + x_n = y_1 + y_2 + ... + y_n$,

we have the conditions

$ x_1 \leq y_1$;
$ x_1 + x_2 \leq y_1 + y_2$;
$ x_1 + x_2 + x_3 \leq y_1 + y_2 + y_3$;
...
generally $ x_1 + x_2 + ... + x_k \leq y_1 + y_2 + ... + y_k$ for any natural k with $ 1 \leq k \leq n - 1$;
and $ x_1 + x_2 + ... + x_n = y_1 + y_2 + ... + y_n$

fulfilled (i. e., the same conditions with all $ \geq$'s replaced by $ \leq$'s), then we say that the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ minorizes the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$. We write this in the form $ \left( x_1;\;x_2;\;...;\;x_n\right) \prec \left( y_1;\;y_2;\;...;\;y_n\right)$. Of course, this is equivalent to $ \left( y_1;\;y_2;\;...;\;y_n\right) \succ \left( x_1;\;x_2;\;...;\;x_n\right)$.

2. Thus we have defined the notions "majorizes" and "minorizes" only for non-increasing arrays (i. e. for arrays satisfying $ x_1 \geq x_2 \geq ... \geq x_n$ and $ y_1 \geq y_2 \geq ... \geq y_n$). Now, assume that $ x_1$, $ x_2$, ..., $ x_n$, $ y_1$, $ y_2$, ..., $ y_n$ are just arbitrary real numbers, without any conditions. Then, let $ \left(X_1;\;X_2;\;...;\;X_n\right)$ be the non-increasing permutation of the array $ \left(x_1;\;x_2;\;...;\;x_n\right)$, i. e. the permutation of the array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ that satisfies $ X_1 \geq X_2 \geq ... \geq X_n$. Similarly, let $ \left(Y_1;\;Y_2;\;...;\;Y_n\right)$ be the non-increasing permutation of the array $ \left(y_1;\;y_2;\;...;\;y_n\right)$, i. e. the permutation of the array $ \left(y_1;\;y_2;\;...;\;y_n\right)$ that satisfies $ Y_1 \geq Y_2 \geq ... \geq Y_n$. The number arrays $ \left(X_1;\;X_2;\;...;\;X_n\right)$ and $ \left(Y_1;\;Y_2;\;...;\;Y_n\right)$ are both non-increasing, and hence we have defined majorization for such arrays.

Then, we say that the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$ if and only if the number array $ \left(X_1;\;X_2;\;...;\;X_n\right)$ majorizes the number array $ \left(Y_1;\;Y_2;\;...;\;Y_n\right)$. In this case, we write $ \left( x_1;\;x_2;\;...;\;x_n\right) \succ \left( y_1;\;y_2;\;...;\;y_n\right)$.

Similarly, we say that the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ minorizes the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$ if and only if the number array $ \left(X_1;\;X_2;\;...;\;X_n\right)$ minorizes the number array $ \left(Y_1;\;Y_2;\;...;\;Y_n\right)$. In this case, we write $ \left( x_1;\;x_2;\;...;\;x_n\right) \prec \left( y_1;\;y_2;\;...;\;y_n\right)$. Again this is equivalent to $ \left( y_1;\;y_2;\;...;\;y_n\right) \succ \left( x_1;\;x_2;\;...;\;x_n\right)$.

3. So we have defined the terms "majorize" and "minorize" for any two number arrays. It should be noted that majorization is a partial order on the set of number arrays, not a total order - i. e., not for every pair of two number arrays $ \left(x_1;\;x_2;\;...;\;x_n\right)$ and $ \left(y_1;\;y_2;\;...;\;y_n\right)$ one can say that either the first one majorizes the second one, or the second one majorizes the first one. It often happens that none of the arrays majorizes or minorizes the other one. But sometimes when you have some special arrays, you can prove that one of them majorizes the other one.

4. Now, the Karamata inequality, also called the Majorization Inequality or the Hardy-Littlewood inequality, states that if $ x_1$, $ x_2$, ..., $ x_n$, $ y_1$, $ y_2$, ..., $ y_n$ are $ 2n$ reals from an interval $ I\subseteq\mathbb{R}$ such that the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$, and $ f: I\to\mathbb{R}$ is any convex function, then

$ f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right) \geq f\left( y_1\right) + f\left( y_2\right) + ... + f\left( y_n\right)$.

If $ f: I\to\mathbb{R}$ is a concave function instead, then we instead have

$ f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right) \leq f\left( y_1\right) + f\left( y_2\right) + ... + f\left( y_n\right)$.

If the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ minorizes the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$ instead of majorizing it, then both inequalities are reversed.

5. The Jensen inequality for real numbers is a special case of the Karamata inequality. In fact, if $ m = \frac {x_1 + x_2 + ... + x_n}{n}$ is the arithmetic mean of the numbers $ x_1$, $ x_2$, ..., $ x_n$, then it is easy to show that the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes the number array $ \left(m;\;m;\;...;\;m\right)$. Hence, the Karamata inequality yields:

If f(x) is any convex function, then

$ f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right) \geq n f\left( m\right)$,

i. e.

$ \frac {f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right)}{n} \geq f\left( m\right)$.

If f(x) is a concave function instead, then we instead have

$ f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right) \leq n f\left( m\right)$,

i. e.

$ \frac {f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right)}{n} \leq f\left( m\right)$.

Of course, this is exactly the Jensen inequality for $ n$ reals.

6. The definition of majorizing and minorizing number arrays given above is somewhat unsatisfying from an intuitive point of view, since it does not help one to imagine how an array majorizing another array looks like. Unfortunately, this is partly inherent to the notion of majorization, which indeed is quite unintuitive. For a - rather facile - visualization of the notion, you can imagine that a number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes a number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$ if the two arrays have the same sum of numbers, but the numbers of the first array are set wider apart than those of the second array, while those of the second array lie closer together. From this intuitive viewpoint, it is clear why the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes the number array $ \left(m;\;m;\;...;\;m\right)$, where $ m = \frac {x_1 + x_2 + ... + x_n}{n}$ is the arithmetic mean of the numbers $ x_1$, $ x_2$, ..., $ x_n$: In fact, the two number arrays have the same sum of elements, but the elements of the second number array lie nearer to each other (they are all equal). Alas, this viewpoint does not help one to really understand what majorization is about.

Well, I know there is more to say. For instance, the Karamata inequality has a kind of converse, but I am not sure how it is formulated, so I leave this to the other MathLinkers more used to inequalities.

Darij
This post has been edited 5 times. Last edited by darij grinberg, Mar 29, 2009, 12:10 AM
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ADMann94
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Where can I find a proof of this?

Alex
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fuzzylogic
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#3 • 2 Y
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ADMann94 wrote:
Where can I find a proof of this?

Alex

Probably in this book.
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ADMann94
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Thanks!

Alex
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Megus
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#5 • 2 Y
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I heard that Karamata inequality is indeed Littlewood-Hardy inequality so which name is correct or maybe both [the question is: which name shall I pull out during the contest ]?
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darij grinberg
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Best refer to it as "Karamata Majorization Inequality". As for Littlewood and Hardy, it's true that they have discovered it independently from Karamata, but I personally have never seen anybody naming it for Littlewood and Hardy.

Darij
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Megus
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#7 • 2 Y
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Ok - I asked beacuse of guys who laughed at me when had heard Karamata :D
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flip2004
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#8 • 3 Y
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Let $ {\mathbf x}=(x_1,x_2,...,x_n)  , {\mathbf y}=(y_1,y_2,...,y_n) $
and $ {\mathbf A}=||a_{i,j}|| $ a $n\times n $ matrix having properties:
1) all $a_{i,j} \ge 0 $ ,
2) $ \sum\limits_{i=1}^na_{i,j}= \sum\limits_{j=1}^{n}a_{i,j}=1 $ for $ i,j=1,2,...,n.$

If $ {\mathbf x} $ is given in $ {\mathbf R}^n $ and $ {\mathbf y}^T=A{\mathbf x}^T $ ( where ${\mathbf x}^T $ denotes the transpose of ${\mathbf x}$) , what we can say about $ {\mathbf y} $ ?
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darij grinberg
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#9 • 6 Y
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Well, we can say that $\left\{ \mathbf{x}\right\} \succ \left\{ \mathbf{y}\right\} $. And conversely, if we have two vectors $\left\{ \mathbf{x}\right\} $ and $\left\{ \mathbf{y}\right\} $ such that $\left\{ \mathbf{x}\right\} \succ \left\{ \mathbf{y}\right\} $, then we can find a matrix $\mathbf{A}$ with the properties 1) and 2) such that $ {\mathbf y}^T=\mathbf{A}{\mathbf x}^T $. This is a result found by Karamata.

Darij
This post has been edited 1 time. Last edited by darij grinberg, Oct 17, 2004, 6:42 PM
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Myth
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Actually, true Jensen inequality is
\[a_1f(x_1)+a_2f(x_2)+\dots+a_nf(x_n)\geq f(a_1x_1+\dots+a_nx_n)\],
where $f$ is convex function, $a_i\in[0,1]$ and $a_1+\dots+a_n=1$.
$a_i=\frac{1}{n}$ is a particular case only.
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flip2004
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Hi all,
here is a very interesting paper related to discussed subject:

[*] A. OSTROWSKI , Sur quelques applications des fonctions convexes et concaves au sens de I. Schur, J.Math.Pures.Appl., (9) 31 (1952) 253-292.

Other possible references :

[1] G.H.Hardy , J.E.Littlewood, G.P\'olya , Some simple inequalities satisfied by convex functions, Messenger Math., 58 (1928/29) 145-152.
[2] J.Karamata , Sur une in\'egalit\'e relative aux fonctions convexes, Publ.Math.Univ. Belgrade 1 (1932) 145-158.
[3] T.Popoviciu , Notes sur les fonctions convexes d'ordre sup\'erieur ,III,Mathematica (Cluj) 16 (1940) 74-86.
[4] T.Popoviciu , Notes sur les fonctions convexes d'ordre sup\'erieur ,IV,Disquisitiones Mathematicae 1(1940) 163-171.
[5] T.Popoviciu , Les fonctions convexes, Actualit\'esci.Ind. No.992,Paris,1945.
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Xixas
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I have a very simple question, which does not let me understand many things. So, can anyone explain me, what functions are called convex and what functions are called concave?
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Myth
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Let $f:[a,b]\to \mathbb{R}$. Then $f$ is called convex function iff
\[k\cdot f(x)+(1-k)\cdot f(y)\geq f(kx+(1-k)y)\quad \forall x,y\in [a,b],\ \forall k\in[0,1].\]

If function $f$ is differentiable then $f$ is convex iff $f''(x)\geq 0$ for all $x\in[a,b]$.
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Xixas
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Another question: which functions are differentable? Can you give an example of non-differentable function?
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Myth
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Take $f(x)=|x|$. It is convex non-differentiable function.
To know what is differentiable functions you need any analysis book.
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perfect_radio
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darij wrote:
Well, I know there is more to say. For instance, the Karamata inequality has a kind of converse, but I am not sure how it is formulated, so I leave this to the other MathLinkers more used to inequalities.

anyone know what is this converse?
This post has been edited 1 time. Last edited by perfect_radio, Apr 23, 2006, 5:11 PM
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Bojan Basic
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I think it is the following (although I see absolutely no use of it):

If $\displaystyle\sum f(x_i)\geqslant\sum f(y_i)$ for every convex function $f$ then $(x)\succ(y)$.

Could somebody post a counter example or (even better) prove this?
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fleeting_guest
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The converse is proved by considering the inequality for:
constant functions (conclusion: vectors $x$ and $y$ have equal number of components),
linear functions (conclusion: $\Sigma x = \Sigma y$), and
functions of the form $\max(0,x-a)$ (conclusion: the majorization conditions).

Any convex $f$ is (on the given finite set of points) equal to a positive linear combination of these 3 types of functions.
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fleeting_guest
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darij grinberg wrote:
Best refer to it as "Karamata Majorization Inequality". As for Littlewood and Hardy, it's true that they have discovered it independently from Karamata, but I personally have never seen anybody naming it for Littlewood and Hardy.

Karamata published later than Hardy,Littlewood and Polya, and in any case majorization was independently discovered many times before and after all these publications. There is a history in Marshall and Olkin's book on majorization. In the Anglo-Saxon countries it is called "majorization inequality" or "Hardy-Littlewood-Polya" majorization inequality".
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darij grinberg
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Bojan Basic wrote:
I think it is the following (although I see absolutely no use of it):

If $\displaystyle\sum f(x_i)\geqslant\sum f(y_i)$ for every convex function $f$ then $(x)\succ(y)$.

There is a stonger version of this: If $\sum f\left(x_i\right)\geq\sum f\left(y_i\right)$ holds for every function f of the form f(x) = |x - u| (with u constant), then $\left(x\right)\succ\left(y\right)$. This is actually Lemma 1 in http://www.mathlinks.ro/Forum/viewtopic.php?t=19097 post #11.

darij
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Peter
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darij grinberg wrote:
5. The Jensen inequality is a special case of the Karamata inequality. In fact, if $\displaystyle m=\frac{x_1+x_2+...+x_n}{n}$ is the arithmetic mean of the numbers $x_1$, $x_2$, ..., $x_n$, then you can easily show (little exercise!) that the number array $\left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes the number array $\left(m;\;m;\;...;\;m\right)$. Hence, the Karamata inequality yields:

If f(x) is any convex function, then

$f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right) \geq n f\left( m\right)$,

i. e.

$\displaystyle \frac{f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right)}{n} \geq f\left( m\right)$.

If f(x) is a concave function instead, then we instead have

$f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right) \leq n f\left( m\right)$,

i. e.

$\displaystyle \frac{f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right)}{n} \leq f\left( m\right)$.

Of course, this is exactly the Jensen inequality.
Myth wrote:
Actually, true Jensen inequality is
\[a_1f(x_1)+a_2f(x_2)+\dots+a_nf(x_n)\geq f(a_1x_1+\dots+a_nx_n)\],
where $f$ is convex function, $a_i\in[0,1]$ and $a_1+\dots+a_n=1$.
$a_i=\frac{1}{n}$ is a particular case only.

Ok, then is there a way to deduce "real" jensen ineq from karamata? :)

It looks like we could extend to rationals and then to reals, but I wonder if someone had a nicer proof, or someone knew sort of a "weighed karamata"?
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k2c901_1
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Is there a continuous or integral analog to this inequality?
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MysticTerminator
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#23 • 2 Y
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Peter VDD wrote:
darij grinberg wrote:
5. The Jensen inequality is a special case of the Karamata inequality. In fact, if $\displaystyle m=\frac{x_1+x_2+...+x_n}{n}$ is the arithmetic mean of the numbers $x_1$, $x_2$, ..., $x_n$, then you can easily show (little exercise!) that the number array $\left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes the number array $\left(m;\;m;\;...;\;m\right)$. Hence, the Karamata inequality yields:

If f(x) is any convex function, then

$f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right) \geq n f\left( m\right)$,

i. e.

$\displaystyle \frac{f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right)}{n} \geq f\left( m\right)$.

If f(x) is a concave function instead, then we instead have

$f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right) \leq n f\left( m\right)$,

i. e.

$\displaystyle \frac{f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right)}{n} \leq f\left( m\right)$.

Of course, this is exactly the Jensen inequality.
Myth wrote:
Actually, true Jensen inequality is
\[a_1f(x_1)+a_2f(x_2)+\dots+a_nf(x_n)\geq f(a_1x_1+\dots+a_nx_n)\],
where $f$ is convex function, $a_i\in[0,1]$ and $a_1+\dots+a_n=1$.
$a_i=\frac{1}{n}$ is a particular case only.

Ok, then is there a way to deduce "real" jensen ineq from karamata? :)

It looks like we could extend to rationals and then to reals, but I wonder if someone had a nicer proof, or someone knew sort of a "weighed karamata"?

I think Darij posted something known as "Fuchs" in theorems & formula section
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lomos_lupin
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Hmm.Guys do we have the Karamata for the productive case to :

Like this:
let $X=(x_1,...,x_n),Y=(y_1,...,y_n)$That $x \succ Y$ and $f$ be a convex function.
then
$f(x_1)*f(x_2)*...*f(x_n) \ge f(y_1)*f(y_2)*...*f(y_n)$

An idea
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Soarer
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this kind of majorization exists, and the inequality you mentioned exists, even for the weighted version I think it exists. But then, you can just take log and it all becomes the normal one.
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Soarer
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#26 • 2 Y
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darij grinberg wrote:
Well, we can say that $\left\{ \mathbf{x}\right\} \succ \left\{ \mathbf{y}\right\} $. And conversely, if we have two vectors $\left\{ \mathbf{x}\right\} $ and $\left\{ \mathbf{y}\right\} $ such that $\left\{ \mathbf{x}\right\} \succ \left\{ \mathbf{y}\right\} $, then we can find a matrix $\mathbf{A}$ with the properties 1) and 2) such that $ {\mathbf y}^T=\mathbf{A}{\mathbf x}^T $. This is a result found by Karamata.

Darij

can you show a proof?
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perfect_radio
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#27 • 2 Y
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i have two questions:

(1) Can Karamata be proven by Jensen?

(2) How do you prove the Fuchs inequality?
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Peter
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#28 • 2 Y
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perfect_radio wrote:
(2) How do you prove the Fuchs inequality?
I wonder about that one too... however I'm afraid that it's not quite elementary.

PS: Can one prove weighted jensen/muirhead from fuchs
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Diarmuid
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#29 • 2 Y
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lomos_lupin wrote:
Hmm.Guys do we have the Karamata for the productive case to :

Like this:
let $X=(x_1,...,x_n),Y=(y_1,...,y_n)$That $x \succ Y$ and $f$ be a convex function.
then
$f(x_1)*f(x_2)*...*f(x_n) \ge f(y_1)*f(y_2)*...*f(y_n)$

My guess (although it's only a guess) would be that you'd need a stronger condition on $f$ for this to hold - probably something like $g(x)=\ln f(x)$ would need to be convex, to convert it back to the additive form.
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DPopov
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#30 • 2 Y
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So is Muirhead simply a special case of Karamata?
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Rzeszut
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k2c901_1 wrote:
Is there a continuous or integral analog to this inequality?
I also wonder if the following is true:
Let $\varphi$ and $\psi$ be two functions defined on interval $\langle a,b\rangle$ satisfying:
1. $\int_{a}^{c}\varphi(t)dt\geq \int_{a}^{c}\psi(t)dt$ for any $c\in \langle a,b\rangle$;
2. $\int_{a}^{b}\varphi(t)dt= \int_{a}^{b}\psi(t)dt$.
Then for any convex function $f$ we have \[\int_{a}^{b}f\left(\varphi(x)\right)\geq \int_{a}^{b}f\left(\psi(x)\right).\] I hope someone can prove it or give a counterexample.
Maybe some majorization theory theorems work also for majorization defined in 1. and 2.? Maybe continuous Muirhead?
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perfect_radio
2607 posts
#32 • 2 Y
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Peter VDD proved here something that reminds me of Karamata for two variables.
Peter VDD wrote:
Theorem.
Let $f(t)$ be a smooth nonnegative function which is convex on $[a,b]$ and let $[x,y]\subset[a,b]$ such that $x+y=a+b$. Then we have
\[\frac{\dsp\int^{b}_{a}f(t) dt}{b-a}\ge \frac{\dsp\int^{y}_{x}f(t) dt}{y-x}.\]

But it's not really what you want. It would be very nice if your conjecture was true :)
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bodan
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#33 • 2 Y
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The inequality Rzeszut proposed exists, and I think even Muirhead for integrals exists. I will post them as soon as I find them.
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bodan
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Look here.
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Rzeszut
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#35 • 2 Y
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bodan wrote:
Look here.
Thanks a lot. :coolspeak:
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sunchips
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#36 • 1 Y
Y by Adventure10
On Olympiad competition, if you use Karamata, do you need justification? or can you just cite it.

For example, I heard that often markers discourage the use of Muirhead.
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amgaa36
5 posts
#37 • 2 Y
Y by Adventure10, Mango247
Is there any example?
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derezin
32 posts
#38 • 2 Y
Y by Adventure10, Mango247
Megus wrote:
Ok - I asked beacuse of guys who laughed at me when had heard Karamata :D
And we will always laugh! :heli:
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BG Yoda
163 posts
#39 • 2 Y
Y by Adventure10, Mango247
Can someone give me links with problems solved by Karamata ineq , please :oops: ?
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quangpbc
533 posts
#40 • 2 Y
Y by Adventure10, Mango247
Eg

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1726203183&t=166411

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1726203183&t=163149

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1726203183&t=160484

Please use Search tool, you can find a lots of ex about Karamata inequality :wink:

http://www.mathlinks.ro/Forum/search.php
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BG Yoda
163 posts
#41 • 1 Y
Y by Adventure10
Thank you very much , quangpbc :thumbup:

i'm new in the forum and because of that i didn't used the search function :blush:
next time i'll know abaut it and i'll use it . thanks for that , too :D
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Fu_Manchu
45 posts
#42 • 2 Y
Y by Adventure10, Mango247
I post a more general version of Karamata's inequality in here.
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