Mixing
Generalized AM-GM Inequality
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I was discussing means with my friend, and I tried to illustrate the concept of geometric mean using the following idea:
Suppose we have two positive quantities $x,y>0$. The simplest geometric object we can make out of those is an $x times y$ rectangle. What if we want a regular rectangle (i.e., a square) that "best approximates this rectangle"?
One possibility is a square of side length $$ell_1 =frac{x+y}{2} ,$$
keeping the perimeter the same at $2x+2y$. Another candidate is
$$ell_2 =sqrt{xy} ,$$
this time keeping the area the same at $xy$.
I then realized I can generalize this idea to higher dimensions: If we have three positive numbers $x,y,z>0$, consider a $x times y times z$ rectangle, and a cube whose side $ell$ is to be decided:
- Keeping the 1-dimensional "length-of-the-skeleton" the same we get
$$4x+4y+4z=12 ell_1 implies ell_1=frac{x+y+z}{3}. $$
- Keeping the 2-dimensional area of the faces the same we get
$$2xy+2xz+2yz=6ell_2^2 implies ell_2=sqrt{frac{xy+xz+yz}{3}}.$$
- Keeping the 3-dimensional volume the same we get
$$x y z =ell_3^3 implies ell_3=sqrt[3]{x y z}.$$
Notice that among the usual arithmetic and geometric means, a different kind of mean has popped up.
This idea can go further, using "$n$-orthotopes" or hyperrectangles, producing $n$ distinct means from any sequence $x_1,dots,x_n$ of positive quantities:
For $1 leq d leq n$ let $e_d(x_1,dots, x_n)$ denote the elementary symmetric polynomial on $n$ symbols of degree $d$. We define
$$ell_d(x_1,dots,x_n) := sqrt[d]{frac{e_d(x_1,dots,x_n)}{binom{n}{d}}}.$$
I have two questions about this:
- Is this concept already known?
- I believe that the AM-GM inequality generalizes to $ell_1 geq ell_2 geq cdots geq ell_n$. Is this correct?
Thank you!
geometry inequality average means
asked Jan 4 at 10:00
user1337user1337
16.6k43391
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add a comment |
$begingroup$
I was discussing means with my friend, and I tried to illustrate the concept of geometric mean using the following idea:
Suppose we have two positive quantities $x,y>0$. The simplest geometric object we can make out of those is an $x times y$ rectangle. What if we want a regular rectangle (i.e., a square) that "best approximates this rectangle"?
One possibility is a square of side length $$ell_1 =frac{x+y}{2} ,$$
keeping the perimeter the same at $2x+2y$. Another candidate is
$$ell_2 =sqrt{xy} ,$$
this time keeping the area the same at $xy$.
I then realized I can generalize this idea to higher dimensions: If we have three positive numbers $x,y,z>0$, consider a $x times y times z$ rectangle, and a cube whose side $ell$ is to be decided:
- Keeping the 1-dimensional "length-of-the-skeleton" the same we get
$$4x+4y+4z=12 ell_1 implies ell_1=frac{x+y+z}{3}. $$
- Keeping the 2-dimensional area of the faces the same we get
$$2xy+2xz+2yz=6ell_2^2 implies ell_2=sqrt{frac{xy+xz+yz}{3}}.$$
- Keeping the 3-dimensional volume the same we get
$$x y z =ell_3^3 implies ell_3=sqrt[3]{x y z}.$$
Notice that among the usual arithmetic and geometric means, a different kind of mean has popped up.
This idea can go further, using "$n$-orthotopes" or hyperrectangles, producing $n$ distinct means from any sequence $x_1,dots,x_n$ of positive quantities:
For $1 leq d leq n$ let $e_d(x_1,dots, x_n)$ denote the elementary symmetric polynomial on $n$ symbols of degree $d$. We define
$$ell_d(x_1,dots,x_n) := sqrt[d]{frac{e_d(x_1,dots,x_n)}{binom{n}{d}}}.$$
I have two questions about this:
- Is this concept already known?
- I believe that the AM-GM inequality generalizes to $ell_1 geq ell_2 geq cdots geq ell_n$. Is this correct?
Thank you!
geometry inequality average means
asked Jan 4 at 10:00
user1337user1337
16.6k43391
$endgroup$
$begingroup$
In 1979 I found a very nice proof of the Maclaurin's inequality. The idea of the proof you can see here: math.stackexchange.com/questions/197955
$endgroup$
– Michael Rozenberg
Jan 4 at 20:10
add a comment |
$begingroup$
I was discussing means with my friend, and I tried to illustrate the concept of geometric mean using the following idea:
Suppose we have two positive quantities $x,y>0$. The simplest geometric object we can make out of those is an $x times y$ rectangle. What if we want a regular rectangle (i.e., a square) that "best approximates this rectangle"?
One possibility is a square of side length $$ell_1 =frac{x+y}{2} ,$$
keeping the perimeter the same at $2x+2y$. Another candidate is
$$ell_2 =sqrt{xy} ,$$
this time keeping the area the same at $xy$.
I then realized I can generalize this idea to higher dimensions: If we have three positive numbers $x,y,z>0$, consider a $x times y times z$ rectangle, and a cube whose side $ell$ is to be decided:
- Keeping the 1-dimensional "length-of-the-skeleton" the same we get
$$4x+4y+4z=12 ell_1 implies ell_1=frac{x+y+z}{3}. $$
- Keeping the 2-dimensional area of the faces the same we get
$$2xy+2xz+2yz=6ell_2^2 implies ell_2=sqrt{frac{xy+xz+yz}{3}}.$$
- Keeping the 3-dimensional volume the same we get
$$x y z =ell_3^3 implies ell_3=sqrt[3]{x y z}.$$
Notice that among the usual arithmetic and geometric means, a different kind of mean has popped up.
This idea can go further, using "$n$-orthotopes" or hyperrectangles, producing $n$ distinct means from any sequence $x_1,dots,x_n$ of positive quantities:
For $1 leq d leq n$ let $e_d(x_1,dots, x_n)$ denote the elementary symmetric polynomial on $n$ symbols of degree $d$. We define
$$ell_d(x_1,dots,x_n) := sqrt[d]{frac{e_d(x_1,dots,x_n)}{binom{n}{d}}}.$$
I have two questions about this:
- Is this concept already known?
- I believe that the AM-GM inequality generalizes to $ell_1 geq ell_2 geq cdots geq ell_n$. Is this correct?
Thank you!
geometry inequality average means
asked Jan 4 at 10:00
user1337user1337
16.6k43391
$endgroup$
I was discussing means with my friend, and I tried to illustrate the concept of geometric mean using the following idea:
Suppose we have two positive quantities $x,y>0$. The simplest geometric object we can make out of those is an $x times y$ rectangle. What if we want a regular rectangle (i.e., a square) that "best approximates this rectangle"?
One possibility is a square of side length $$ell_1 =frac{x+y}{2} ,$$
keeping the perimeter the same at $2x+2y$. Another candidate is
$$ell_2 =sqrt{xy} ,$$
this time keeping the area the same at $xy$.
I then realized I can generalize this idea to higher dimensions: If we have three positive numbers $x,y,z>0$, consider a $x times y times z$ rectangle, and a cube whose side $ell$ is to be decided:
- Keeping the 1-dimensional "length-of-the-skeleton" the same we get
$$4x+4y+4z=12 ell_1 implies ell_1=frac{x+y+z}{3}. $$
- Keeping the 2-dimensional area of the faces the same we get
$$2xy+2xz+2yz=6ell_2^2 implies ell_2=sqrt{frac{xy+xz+yz}{3}}.$$
- Keeping the 3-dimensional volume the same we get
$$x y z =ell_3^3 implies ell_3=sqrt[3]{x y z}.$$
Notice that among the usual arithmetic and geometric means, a different kind of mean has popped up.
This idea can go further, using "$n$-orthotopes" or hyperrectangles, producing $n$ distinct means from any sequence $x_1,dots,x_n$ of positive quantities:
For $1 leq d leq n$ let $e_d(x_1,dots, x_n)$ denote the elementary symmetric polynomial on $n$ symbols of degree $d$. We define
$$ell_d(x_1,dots,x_n) := sqrt[d]{frac{e_d(x_1,dots,x_n)}{binom{n}{d}}}.$$
I have two questions about this:
- Is this concept already known?
- I believe that the AM-GM inequality generalizes to $ell_1 geq ell_2 geq cdots geq ell_n$. Is this correct?
Thank you!
geometry inequality average means
geometry inequality average means
asked Jan 4 at 10:00
user1337user1337
16.6k43391
asked Jan 4 at 10:00
user1337user1337
16.6k43391
asked Jan 4 at 10:00
user1337user1337
16.6k43391
asked Jan 4 at 10:00
user1337user1337
16.6k43391
asked Jan 4 at 10:00
user1337user1337
16.6k43391
16.6k43391
$begingroup$
In 1979 I found a very nice proof of the Maclaurin's inequality. The idea of the proof you can see here: math.stackexchange.com/questions/197955
$endgroup$
– Michael Rozenberg
Jan 4 at 20:10
add a comment |
$begingroup$
In 1979 I found a very nice proof of the Maclaurin's inequality. The idea of the proof you can see here: math.stackexchange.com/questions/197955
$endgroup$
– Michael Rozenberg
Jan 4 at 20:10
$begingroup$
In 1979 I found a very nice proof of the Maclaurin's inequality. The idea of the proof you can see here: math.stackexchange.com/questions/197955
$endgroup$
– Michael Rozenberg
Jan 4 at 20:10
$begingroup$
In 1979 I found a very nice proof of the Maclaurin's inequality. The idea of the proof you can see here: math.stackexchange.com/questions/197955
$endgroup$
– Michael Rozenberg
Jan 4 at 20:10
add a comment |
1 Answer
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Yes, that relationship between the elementary symmetric polynomials holds, it is known as Maclaurin's inequality, and a consequence of Newton's inequalities.
answered Jan 4 at 10:15
Martin RMartin R
27.8k33255
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add a comment |
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$begingroup$
Yes, that relationship between the elementary symmetric polynomials holds, it is known as Maclaurin's inequality, and a consequence of Newton's inequalities.
answered Jan 4 at 10:15
Martin RMartin R
27.8k33255
$endgroup$
add a comment |
$begingroup$
Yes, that relationship between the elementary symmetric polynomials holds, it is known as Maclaurin's inequality, and a consequence of Newton's inequalities.
answered Jan 4 at 10:15
Martin RMartin R
27.8k33255
$endgroup$
add a comment |
$begingroup$
Yes, that relationship between the elementary symmetric polynomials holds, it is known as Maclaurin's inequality, and a consequence of Newton's inequalities.
answered Jan 4 at 10:15
Martin RMartin R
27.8k33255
$endgroup$
Yes, that relationship between the elementary symmetric polynomials holds, it is known as Maclaurin's inequality, and a consequence of Newton's inequalities.
answered Jan 4 at 10:15
Martin RMartin R
27.8k33255
answered Jan 4 at 10:15
Martin RMartin R
27.8k33255
answered Jan 4 at 10:15
Martin RMartin R
27.8k33255
answered Jan 4 at 10:15
Martin RMartin R
27.8k33255
27.8k33255
add a comment |
add a comment |
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$begingroup$
In 1979 I found a very nice proof of the Maclaurin's inequality. The idea of the proof you can see here: math.stackexchange.com/questions/197955
$endgroup$
– Michael Rozenberg
Jan 4 at 20:10