Generalized AM-GM Inequality
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 yesterday
user1337
16.6k43391
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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 yesterday
user1337
16.6k43391
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
– Michael Rozenberg
22 hours ago
add a comment |
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 yesterday
user1337
16.6k43391
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 yesterday
user1337
16.6k43391
asked yesterday
user1337
16.6k43391
asked yesterday
user1337
16.6k43391
asked yesterday
user1337
16.6k43391
asked yesterday
user1337
16.6k43391
16.6k43391
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
– Michael Rozenberg
22 hours ago
add a comment |
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
– Michael Rozenberg
22 hours ago
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
– Michael Rozenberg
22 hours ago
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
– Michael Rozenberg
22 hours ago
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 yesterday
Martin R
27.3k33254
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes, that relationship between the elementary symmetric polynomials holds, it is known as Maclaurin's inequality, and a consequence of Newton's inequalities.
answered yesterday
Martin R
27.3k33254
add a comment |
Yes, that relationship between the elementary symmetric polynomials holds, it is known as Maclaurin's inequality, and a consequence of Newton's inequalities.
answered yesterday
Martin R
27.3k33254
add a comment |
Yes, that relationship between the elementary symmetric polynomials holds, it is known as Maclaurin's inequality, and a consequence of Newton's inequalities.
answered yesterday
Martin R
27.3k33254
Yes, that relationship between the elementary symmetric polynomials holds, it is known as Maclaurin's inequality, and a consequence of Newton's inequalities.
answered yesterday
Martin R
27.3k33254
answered yesterday
Martin R
27.3k33254
answered yesterday
Martin R
27.3k33254
answered yesterday
Martin R
27.3k33254
27.3k33254
add a comment |
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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
– Michael Rozenberg
22 hours ago