GM-AM Inequality Proof
Clash Royale CLAN TAG#URR8PPP
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I encountered this problem in Kuratowski's intro to calculus. I'd like help in the direction of the hint provided in the book, please. In the hint, I don't understand how the geometric mean is greater than $G$, or where it's even headed. Thank you.
Denote by $A$ and $G$ the arithmetic and geometric means of the numbers $a_1,a_2,...,a_n,$ respectively, i.e.,
$$A=fraca_1+a_2+...a_nn, G=sqrt[leftroot-2uproot2n]a_1cdot a_2cdot ...cdot a_n.$$
Prove that if the numbers $a_1,a_2,...,a_n$ are positive, then $Gle A$.
Hint: Precede the proof by the remark that if $a_1<A<a_2$ then the arithmetic mean of the numbers $A,a_1+a_2-A,a_3,...,a_n$ is $A$ and the geometric mean of these numbers is $>G$.
By following the hint I get that $$A(a_1+a_2-A)>a_1a_2$$
How is this true?
calculus real-analysis
asked Aug 30 at 4:19
Alex D
444217
add a comment |Â
up vote
5
down vote
favorite
I encountered this problem in Kuratowski's intro to calculus. I'd like help in the direction of the hint provided in the book, please. In the hint, I don't understand how the geometric mean is greater than $G$, or where it's even headed. Thank you.
Denote by $A$ and $G$ the arithmetic and geometric means of the numbers $a_1,a_2,...,a_n,$ respectively, i.e.,
$$A=fraca_1+a_2+...a_nn, G=sqrt[leftroot-2uproot2n]a_1cdot a_2cdot ...cdot a_n.$$
Prove that if the numbers $a_1,a_2,...,a_n$ are positive, then $Gle A$.
Hint: Precede the proof by the remark that if $a_1<A<a_2$ then the arithmetic mean of the numbers $A,a_1+a_2-A,a_3,...,a_n$ is $A$ and the geometric mean of these numbers is $>G$.
By following the hint I get that $$A(a_1+a_2-A)>a_1a_2$$
How is this true?
calculus real-analysis
asked Aug 30 at 4:19
Alex D
444217
You want to prove in this way only or you want any proof en.wikipedia.org/wiki/… has good proof for this
– Deepesh Meena
Aug 30 at 4:32
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I encountered this problem in Kuratowski's intro to calculus. I'd like help in the direction of the hint provided in the book, please. In the hint, I don't understand how the geometric mean is greater than $G$, or where it's even headed. Thank you.
Denote by $A$ and $G$ the arithmetic and geometric means of the numbers $a_1,a_2,...,a_n,$ respectively, i.e.,
$$A=fraca_1+a_2+...a_nn, G=sqrt[leftroot-2uproot2n]a_1cdot a_2cdot ...cdot a_n.$$
Prove that if the numbers $a_1,a_2,...,a_n$ are positive, then $Gle A$.
Hint: Precede the proof by the remark that if $a_1<A<a_2$ then the arithmetic mean of the numbers $A,a_1+a_2-A,a_3,...,a_n$ is $A$ and the geometric mean of these numbers is $>G$.
By following the hint I get that $$A(a_1+a_2-A)>a_1a_2$$
How is this true?
calculus real-analysis
asked Aug 30 at 4:19
Alex D
444217
I encountered this problem in Kuratowski's intro to calculus. I'd like help in the direction of the hint provided in the book, please. In the hint, I don't understand how the geometric mean is greater than $G$, or where it's even headed. Thank you.
Denote by $A$ and $G$ the arithmetic and geometric means of the numbers $a_1,a_2,...,a_n,$ respectively, i.e.,
$$A=fraca_1+a_2+...a_nn, G=sqrt[leftroot-2uproot2n]a_1cdot a_2cdot ...cdot a_n.$$
Prove that if the numbers $a_1,a_2,...,a_n$ are positive, then $Gle A$.
Hint: Precede the proof by the remark that if $a_1<A<a_2$ then the arithmetic mean of the numbers $A,a_1+a_2-A,a_3,...,a_n$ is $A$ and the geometric mean of these numbers is $>G$.
By following the hint I get that $$A(a_1+a_2-A)>a_1a_2$$
How is this true?
calculus real-analysis
calculus real-analysis
asked Aug 30 at 4:19
Alex D
444217
asked Aug 30 at 4:19
Alex D
444217
edited 2 days ago
asked Aug 30 at 4:19
Alex D
444217
asked Aug 30 at 4:19
Alex D
444217
asked Aug 30 at 4:19
Alex D
444217
444217
You want to prove in this way only or you want any proof en.wikipedia.org/wiki/… has good proof for this
– Deepesh Meena
Aug 30 at 4:32
add a comment |Â
You want to prove in this way only or you want any proof en.wikipedia.org/wiki/… has good proof for this
– Deepesh Meena
Aug 30 at 4:32
You want to prove in this way only or you want any proof en.wikipedia.org/wiki/… has good proof for this
– Deepesh Meena
Aug 30 at 4:32
You want to prove in this way only or you want any proof en.wikipedia.org/wiki/… has good proof for this
– Deepesh Meena
Aug 30 at 4:32
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Here's where the hint is headed. If the numbers in your list are not all equal, then there exist two numbers, call them $a_1$ and $a_2$, such that $A$ lies strictly between them. Create a new list from the original list by replacing $a_1$ with $A$, and replacing $a_2$ with $a_1+a_2-A$. Then, according the the hint (which you've proved via @dxiv's hint), the new list has the same arithmetic mean as, but larger geometric mean than, the original. What next? You can repeat this procedure until you obtain a list consisting of all $A$. What can you say about the arithmetic mean and geometric mean of this final list, and how does these relate to the original list?
answered Aug 30 at 4:56
grand_chat
18.3k11122
Thank you, I was able to figure it out with your hints. I have limited internet access, so sorry for the delayed response.
– Alex D
2 days ago
add a comment |Â
up vote
3
down vote
By following the hint I get that: $quad A(a_1+a_2-A)>a_1a_2$
Next hint: Â that's equivalent to $,(A-a_1)(A-a_2) lt 0,$.
answered Aug 30 at 4:35
dxiv
55.8k64798
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Here's where the hint is headed. If the numbers in your list are not all equal, then there exist two numbers, call them $a_1$ and $a_2$, such that $A$ lies strictly between them. Create a new list from the original list by replacing $a_1$ with $A$, and replacing $a_2$ with $a_1+a_2-A$. Then, according the the hint (which you've proved via @dxiv's hint), the new list has the same arithmetic mean as, but larger geometric mean than, the original. What next? You can repeat this procedure until you obtain a list consisting of all $A$. What can you say about the arithmetic mean and geometric mean of this final list, and how does these relate to the original list?
answered Aug 30 at 4:56
grand_chat
18.3k11122
Thank you, I was able to figure it out with your hints. I have limited internet access, so sorry for the delayed response.
– Alex D
2 days ago
add a comment |Â
up vote
3
down vote
accepted
Here's where the hint is headed. If the numbers in your list are not all equal, then there exist two numbers, call them $a_1$ and $a_2$, such that $A$ lies strictly between them. Create a new list from the original list by replacing $a_1$ with $A$, and replacing $a_2$ with $a_1+a_2-A$. Then, according the the hint (which you've proved via @dxiv's hint), the new list has the same arithmetic mean as, but larger geometric mean than, the original. What next? You can repeat this procedure until you obtain a list consisting of all $A$. What can you say about the arithmetic mean and geometric mean of this final list, and how does these relate to the original list?
answered Aug 30 at 4:56
grand_chat
18.3k11122
Thank you, I was able to figure it out with your hints. I have limited internet access, so sorry for the delayed response.
– Alex D
2 days ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Here's where the hint is headed. If the numbers in your list are not all equal, then there exist two numbers, call them $a_1$ and $a_2$, such that $A$ lies strictly between them. Create a new list from the original list by replacing $a_1$ with $A$, and replacing $a_2$ with $a_1+a_2-A$. Then, according the the hint (which you've proved via @dxiv's hint), the new list has the same arithmetic mean as, but larger geometric mean than, the original. What next? You can repeat this procedure until you obtain a list consisting of all $A$. What can you say about the arithmetic mean and geometric mean of this final list, and how does these relate to the original list?
answered Aug 30 at 4:56
grand_chat
18.3k11122
Here's where the hint is headed. If the numbers in your list are not all equal, then there exist two numbers, call them $a_1$ and $a_2$, such that $A$ lies strictly between them. Create a new list from the original list by replacing $a_1$ with $A$, and replacing $a_2$ with $a_1+a_2-A$. Then, according the the hint (which you've proved via @dxiv's hint), the new list has the same arithmetic mean as, but larger geometric mean than, the original. What next? You can repeat this procedure until you obtain a list consisting of all $A$. What can you say about the arithmetic mean and geometric mean of this final list, and how does these relate to the original list?
answered Aug 30 at 4:56
grand_chat
18.3k11122
answered Aug 30 at 4:56
grand_chat
18.3k11122
answered Aug 30 at 4:56
grand_chat
18.3k11122
answered Aug 30 at 4:56
grand_chat
18.3k11122
18.3k11122
Thank you, I was able to figure it out with your hints. I have limited internet access, so sorry for the delayed response.
– Alex D
2 days ago
add a comment |Â
Thank you, I was able to figure it out with your hints. I have limited internet access, so sorry for the delayed response.
– Alex D
2 days ago
Thank you, I was able to figure it out with your hints. I have limited internet access, so sorry for the delayed response.
– Alex D
2 days ago
Thank you, I was able to figure it out with your hints. I have limited internet access, so sorry for the delayed response.
– Alex D
2 days ago
add a comment |Â
up vote
3
down vote
By following the hint I get that: $quad A(a_1+a_2-A)>a_1a_2$
Next hint: Â that's equivalent to $,(A-a_1)(A-a_2) lt 0,$.
answered Aug 30 at 4:35
dxiv
55.8k64798
add a comment |Â
up vote
3
down vote
By following the hint I get that: $quad A(a_1+a_2-A)>a_1a_2$
Next hint: Â that's equivalent to $,(A-a_1)(A-a_2) lt 0,$.
answered Aug 30 at 4:35
dxiv
55.8k64798
add a comment |Â
up vote
3
down vote
up vote
3
down vote
By following the hint I get that: $quad A(a_1+a_2-A)>a_1a_2$
Next hint: Â that's equivalent to $,(A-a_1)(A-a_2) lt 0,$.
answered Aug 30 at 4:35
dxiv
55.8k64798
By following the hint I get that: $quad A(a_1+a_2-A)>a_1a_2$
Next hint: Â that's equivalent to $,(A-a_1)(A-a_2) lt 0,$.
answered Aug 30 at 4:35
dxiv
55.8k64798
answered Aug 30 at 4:35
dxiv
55.8k64798
answered Aug 30 at 4:35
dxiv
55.8k64798
answered Aug 30 at 4:35
dxiv
55.8k64798
55.8k64798
add a comment |Â
add a comment |Â
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You want to prove in this way only or you want any proof en.wikipedia.org/wiki/… has good proof for this
– Deepesh Meena
Aug 30 at 4:32