Akerberg's Refinement of AM-GM: Motivation of proofShowing $sumlimits^N_n=1left(prodlimits_i=1^n b_i right)^frac1nlesumlimits^N_n=1left(prodlimits_i=1^n a_i right)^frac1n$?A symmetric inequalityHow to understand Cauchy's proof of AM-GM inequality(the last step)Use the arithmetic-geometric inequality for this list to deduce the arithmetic-geometric inequality for $n$.What is the motivation of this inequality?Prove $sumlimits_textcycfracaa+(n-1)bgeq 1$Proving an inequality via the Cauchy-Schwarz InequalityProving Cauchy-Schwarz with Arithmetic Geometric mean$L^1$ inequality between ordered real numbers implies $L^2$ norm inequalityFinding equality of inequality via Cauchy-Schwarz
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Akerberg's Refinement of AM-GM: Motivation of proof
Showing $sumlimits^N_n=1left(prodlimits_i=1^n b_i right)^frac1nlesumlimits^N_n=1left(prodlimits_i=1^n a_i right)^frac1n$?A symmetric inequalityHow to understand Cauchy's proof of AM-GM inequality(the last step)Use the arithmetic-geometric inequality for this list to deduce the arithmetic-geometric inequality for $n$.What is the motivation of this inequality?Prove $sumlimits_textcycfracaa+(n-1)bgeq 1$Proving an inequality via the Cauchy-Schwarz InequalityProving Cauchy-Schwarz with Arithmetic Geometric mean$L^1$ inequality between ordered real numbers implies $L^2$ norm inequalityFinding equality of inequality via Cauchy-Schwarz
$begingroup$
In Steele's Cauchy Schwarz Master Class Exercise 2.10 is about Akerberg's Refinement of AM-GM. (It's a refinement because AM-GM follows by iteration.)
Problem. For $a_1, dots, a_n geq 0$ and $n geq 2$ prove $$ a_n left(fraca_1+dots+a_n-1n-1 right)^n-1 leq left(fraca_1+dots+a_nn right)^n. $$ The book proposes to use (a consequence of Bernoulli) $$y left(n-y^n-1 right) = ny-y^n leq n-1 quad textfor quad y geq 0.$$ By setting $$y^n-1 = fraca_noverlinea quad textwith quad overlinea = fraca_1+dots+a_nn $$ we are immediately done.
However, how is $y left(n-y^n-1 right) leq n-1$ related to the given problem? Where is the motivation to look at that? How would one think of the choice of $y$?
inequality a.m.-g.m.-inequality
asked Mar 10 at 20:50
KezerKezer
1,458621
$endgroup$
add a comment |
$begingroup$
In Steele's Cauchy Schwarz Master Class Exercise 2.10 is about Akerberg's Refinement of AM-GM. (It's a refinement because AM-GM follows by iteration.)
Problem. For $a_1, dots, a_n geq 0$ and $n geq 2$ prove $$ a_n left(fraca_1+dots+a_n-1n-1 right)^n-1 leq left(fraca_1+dots+a_nn right)^n. $$ The book proposes to use (a consequence of Bernoulli) $$y left(n-y^n-1 right) = ny-y^n leq n-1 quad textfor quad y geq 0.$$ By setting $$y^n-1 = fraca_noverlinea quad textwith quad overlinea = fraca_1+dots+a_nn $$ we are immediately done.
However, how is $y left(n-y^n-1 right) leq n-1$ related to the given problem? Where is the motivation to look at that? How would one think of the choice of $y$?
inequality a.m.-g.m.-inequality
asked Mar 10 at 20:50
KezerKezer
1,458621
$endgroup$
$begingroup$
Perhaps also interesting: There is a proof by the weighted AM-GM. Take the $n$-th root. Then, the candidates for the inequality are already written out.
$endgroup$
– Kezer
Mar 10 at 20:52
add a comment |
$begingroup$
In Steele's Cauchy Schwarz Master Class Exercise 2.10 is about Akerberg's Refinement of AM-GM. (It's a refinement because AM-GM follows by iteration.)
Problem. For $a_1, dots, a_n geq 0$ and $n geq 2$ prove $$ a_n left(fraca_1+dots+a_n-1n-1 right)^n-1 leq left(fraca_1+dots+a_nn right)^n. $$ The book proposes to use (a consequence of Bernoulli) $$y left(n-y^n-1 right) = ny-y^n leq n-1 quad textfor quad y geq 0.$$ By setting $$y^n-1 = fraca_noverlinea quad textwith quad overlinea = fraca_1+dots+a_nn $$ we are immediately done.
However, how is $y left(n-y^n-1 right) leq n-1$ related to the given problem? Where is the motivation to look at that? How would one think of the choice of $y$?
inequality a.m.-g.m.-inequality
asked Mar 10 at 20:50
KezerKezer
1,458621
$endgroup$
In Steele's Cauchy Schwarz Master Class Exercise 2.10 is about Akerberg's Refinement of AM-GM. (It's a refinement because AM-GM follows by iteration.)
Problem. For $a_1, dots, a_n geq 0$ and $n geq 2$ prove $$ a_n left(fraca_1+dots+a_n-1n-1 right)^n-1 leq left(fraca_1+dots+a_nn right)^n. $$ The book proposes to use (a consequence of Bernoulli) $$y left(n-y^n-1 right) = ny-y^n leq n-1 quad textfor quad y geq 0.$$ By setting $$y^n-1 = fraca_noverlinea quad textwith quad overlinea = fraca_1+dots+a_nn $$ we are immediately done.
However, how is $y left(n-y^n-1 right) leq n-1$ related to the given problem? Where is the motivation to look at that? How would one think of the choice of $y$?
inequality a.m.-g.m.-inequality
inequality a.m.-g.m.-inequality
asked Mar 10 at 20:50
KezerKezer
1,458621
asked Mar 10 at 20:50
KezerKezer
1,458621
asked Mar 10 at 20:50
KezerKezer
1,458621
asked Mar 10 at 20:50
KezerKezer
1,458621
asked Mar 10 at 20:50
KezerKezer
1,458621
1,458621
$begingroup$
Perhaps also interesting: There is a proof by the weighted AM-GM. Take the $n$-th root. Then, the candidates for the inequality are already written out.
$endgroup$
– Kezer
Mar 10 at 20:52
add a comment |
$begingroup$
Perhaps also interesting: There is a proof by the weighted AM-GM. Take the $n$-th root. Then, the candidates for the inequality are already written out.
$endgroup$
– Kezer
Mar 10 at 20:52
$begingroup$
Perhaps also interesting: There is a proof by the weighted AM-GM. Take the $n$-th root. Then, the candidates for the inequality are already written out.
$endgroup$
– Kezer
Mar 10 at 20:52
$begingroup$
Perhaps also interesting: There is a proof by the weighted AM-GM. Take the $n$-th root. Then, the candidates for the inequality are already written out.
$endgroup$
– Kezer
Mar 10 at 20:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Motivation is linked to possible thought process - and of course multiple processes could motivate the same substitution / approach.
For e.g. if one were to observe the inequality to prove is homogeneous (i.e. invariant to scale), one possibility is to simply scale s.t. $a_1+a_2+cdots+a_n-1 = n-1$, so that the inequality to prove is simply
$$a_n leqslant left(fracn-1+a_nn right)^n = left(1+fraca_n-1n right)^n$$
which of course follows directly from Bernoulli's inequality. The scaling $a_1+a_2+cdots a_n = n$ would also lead to a similar direct application.
In the case you mention, instead of scaling, perhaps the author noted that we can reduce the relevant variables $a_i$ to two (as $a_1+a_2 + cdots + a_n-1$ always appears together), by setting the substitution in $overline a$ and $y$, the inequality to prove reduces to $y(n-y^n-1) leqslant n-1$, which in turn is a simple application of Bernoulli. When writing out, the order is reversed, as the implications are clearer then - unfortunately motivation is sometimes not obvious afterwards.
answered 2 days ago
MacavityMacavity
35.6k52554
$endgroup$
2
$begingroup$
Very nice answer, thanks! I somehow got confused and thought the inequality wasn‘t homogeneous in my thought process (which it clearly is, though). I also like your motivation to think of the problem as having $2$ relevant variables! And yes indeed, proofs are often not written in the same way as they are found (which is why I always try to find the motivation, as that is often the real gem).
$endgroup$
– Kezer
2 days ago
add a comment |
$begingroup$
By AM-GM
$$left(fraca_1+a_2+...+a_nnright)^n=left(fracfraca_1+a_2+...+a_n-1n-1cdot(n-1)+a_nnright)^ngeq$$
$$geqleft(sqrt[n]left(fraca_1+a_2+...+a_n-1n-1right)^n-1 a_nright)^n=a_nleft(fraca_1+a_2+...+a_n-1n-1right)^n-1.$$
There is also the following proof by Bernoulli.
Let $overlinea=fraca_1+...+a_n-1n-1.$
Thus, $$left(fraca_1+a_2+...+a_nnright)^n=left(fracfraca_1+a_2+...+a_n-1n-1cdot(n-1)+a_nnright)^n=$$
$$=left(frac(n-1)overlinea+a_nnright)^n=overlinea^nleft(1+fraca_nnoverlinea-frac1nright)^ngeqoverlinea^nleft(1+nleft(fraca_nnoverlinea-frac1nright)right)=a_noverlinea^n-1$$
answered Mar 10 at 21:28
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
And again...Brilliant!
$endgroup$
– Dr. Mathva
Mar 10 at 21:31
3
$begingroup$
This does not answer my question and is exactly the proof that I mentioned in the comments.
$endgroup$
– Kezer
Mar 10 at 21:37
1
$begingroup$
The proof by Bernoulli that you edited in is exactly the proof that I‘m talking about in the thread. Again, you do not answer my question. I know and appreciate that you are a master at inequalities but nonetheless, your answer unfortunately does not help me here.
$endgroup$
– Kezer
2 days ago
$begingroup$
@Kezer I just don't know English. :) But I tried to help you.
$endgroup$
– Michael Rozenberg
2 days ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Motivation is linked to possible thought process - and of course multiple processes could motivate the same substitution / approach.
For e.g. if one were to observe the inequality to prove is homogeneous (i.e. invariant to scale), one possibility is to simply scale s.t. $a_1+a_2+cdots+a_n-1 = n-1$, so that the inequality to prove is simply
$$a_n leqslant left(fracn-1+a_nn right)^n = left(1+fraca_n-1n right)^n$$
which of course follows directly from Bernoulli's inequality. The scaling $a_1+a_2+cdots a_n = n$ would also lead to a similar direct application.
In the case you mention, instead of scaling, perhaps the author noted that we can reduce the relevant variables $a_i$ to two (as $a_1+a_2 + cdots + a_n-1$ always appears together), by setting the substitution in $overline a$ and $y$, the inequality to prove reduces to $y(n-y^n-1) leqslant n-1$, which in turn is a simple application of Bernoulli. When writing out, the order is reversed, as the implications are clearer then - unfortunately motivation is sometimes not obvious afterwards.
answered 2 days ago
MacavityMacavity
35.6k52554
$endgroup$
2
$begingroup$
Very nice answer, thanks! I somehow got confused and thought the inequality wasn‘t homogeneous in my thought process (which it clearly is, though). I also like your motivation to think of the problem as having $2$ relevant variables! And yes indeed, proofs are often not written in the same way as they are found (which is why I always try to find the motivation, as that is often the real gem).
$endgroup$
– Kezer
2 days ago
add a comment |
$begingroup$
Motivation is linked to possible thought process - and of course multiple processes could motivate the same substitution / approach.
For e.g. if one were to observe the inequality to prove is homogeneous (i.e. invariant to scale), one possibility is to simply scale s.t. $a_1+a_2+cdots+a_n-1 = n-1$, so that the inequality to prove is simply
$$a_n leqslant left(fracn-1+a_nn right)^n = left(1+fraca_n-1n right)^n$$
which of course follows directly from Bernoulli's inequality. The scaling $a_1+a_2+cdots a_n = n$ would also lead to a similar direct application.
In the case you mention, instead of scaling, perhaps the author noted that we can reduce the relevant variables $a_i$ to two (as $a_1+a_2 + cdots + a_n-1$ always appears together), by setting the substitution in $overline a$ and $y$, the inequality to prove reduces to $y(n-y^n-1) leqslant n-1$, which in turn is a simple application of Bernoulli. When writing out, the order is reversed, as the implications are clearer then - unfortunately motivation is sometimes not obvious afterwards.
answered 2 days ago
MacavityMacavity
35.6k52554
$endgroup$
2
$begingroup$
Very nice answer, thanks! I somehow got confused and thought the inequality wasn‘t homogeneous in my thought process (which it clearly is, though). I also like your motivation to think of the problem as having $2$ relevant variables! And yes indeed, proofs are often not written in the same way as they are found (which is why I always try to find the motivation, as that is often the real gem).
$endgroup$
– Kezer
2 days ago
add a comment |
$begingroup$
Motivation is linked to possible thought process - and of course multiple processes could motivate the same substitution / approach.
For e.g. if one were to observe the inequality to prove is homogeneous (i.e. invariant to scale), one possibility is to simply scale s.t. $a_1+a_2+cdots+a_n-1 = n-1$, so that the inequality to prove is simply
$$a_n leqslant left(fracn-1+a_nn right)^n = left(1+fraca_n-1n right)^n$$
which of course follows directly from Bernoulli's inequality. The scaling $a_1+a_2+cdots a_n = n$ would also lead to a similar direct application.
In the case you mention, instead of scaling, perhaps the author noted that we can reduce the relevant variables $a_i$ to two (as $a_1+a_2 + cdots + a_n-1$ always appears together), by setting the substitution in $overline a$ and $y$, the inequality to prove reduces to $y(n-y^n-1) leqslant n-1$, which in turn is a simple application of Bernoulli. When writing out, the order is reversed, as the implications are clearer then - unfortunately motivation is sometimes not obvious afterwards.
answered 2 days ago
MacavityMacavity
35.6k52554
$endgroup$
Motivation is linked to possible thought process - and of course multiple processes could motivate the same substitution / approach.
For e.g. if one were to observe the inequality to prove is homogeneous (i.e. invariant to scale), one possibility is to simply scale s.t. $a_1+a_2+cdots+a_n-1 = n-1$, so that the inequality to prove is simply
$$a_n leqslant left(fracn-1+a_nn right)^n = left(1+fraca_n-1n right)^n$$
which of course follows directly from Bernoulli's inequality. The scaling $a_1+a_2+cdots a_n = n$ would also lead to a similar direct application.
In the case you mention, instead of scaling, perhaps the author noted that we can reduce the relevant variables $a_i$ to two (as $a_1+a_2 + cdots + a_n-1$ always appears together), by setting the substitution in $overline a$ and $y$, the inequality to prove reduces to $y(n-y^n-1) leqslant n-1$, which in turn is a simple application of Bernoulli. When writing out, the order is reversed, as the implications are clearer then - unfortunately motivation is sometimes not obvious afterwards.
answered 2 days ago
MacavityMacavity
35.6k52554
answered 2 days ago
MacavityMacavity
35.6k52554
answered 2 days ago
MacavityMacavity
35.6k52554
answered 2 days ago
MacavityMacavity
35.6k52554
35.6k52554
2
$begingroup$
Very nice answer, thanks! I somehow got confused and thought the inequality wasn‘t homogeneous in my thought process (which it clearly is, though). I also like your motivation to think of the problem as having $2$ relevant variables! And yes indeed, proofs are often not written in the same way as they are found (which is why I always try to find the motivation, as that is often the real gem).
$endgroup$
– Kezer
2 days ago
add a comment |
2
$begingroup$
Very nice answer, thanks! I somehow got confused and thought the inequality wasn‘t homogeneous in my thought process (which it clearly is, though). I also like your motivation to think of the problem as having $2$ relevant variables! And yes indeed, proofs are often not written in the same way as they are found (which is why I always try to find the motivation, as that is often the real gem).
$endgroup$
– Kezer
2 days ago
2
2
$begingroup$
Very nice answer, thanks! I somehow got confused and thought the inequality wasn‘t homogeneous in my thought process (which it clearly is, though). I also like your motivation to think of the problem as having $2$ relevant variables! And yes indeed, proofs are often not written in the same way as they are found (which is why I always try to find the motivation, as that is often the real gem).
$endgroup$
– Kezer
2 days ago
$begingroup$
Very nice answer, thanks! I somehow got confused and thought the inequality wasn‘t homogeneous in my thought process (which it clearly is, though). I also like your motivation to think of the problem as having $2$ relevant variables! And yes indeed, proofs are often not written in the same way as they are found (which is why I always try to find the motivation, as that is often the real gem).
$endgroup$
– Kezer
2 days ago
add a comment |
$begingroup$
By AM-GM
$$left(fraca_1+a_2+...+a_nnright)^n=left(fracfraca_1+a_2+...+a_n-1n-1cdot(n-1)+a_nnright)^ngeq$$
$$geqleft(sqrt[n]left(fraca_1+a_2+...+a_n-1n-1right)^n-1 a_nright)^n=a_nleft(fraca_1+a_2+...+a_n-1n-1right)^n-1.$$
There is also the following proof by Bernoulli.
Let $overlinea=fraca_1+...+a_n-1n-1.$
Thus, $$left(fraca_1+a_2+...+a_nnright)^n=left(fracfraca_1+a_2+...+a_n-1n-1cdot(n-1)+a_nnright)^n=$$
$$=left(frac(n-1)overlinea+a_nnright)^n=overlinea^nleft(1+fraca_nnoverlinea-frac1nright)^ngeqoverlinea^nleft(1+nleft(fraca_nnoverlinea-frac1nright)right)=a_noverlinea^n-1$$
answered Mar 10 at 21:28
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
And again...Brilliant!
$endgroup$
– Dr. Mathva
Mar 10 at 21:31
3
$begingroup$
This does not answer my question and is exactly the proof that I mentioned in the comments.
$endgroup$
– Kezer
Mar 10 at 21:37
1
$begingroup$
The proof by Bernoulli that you edited in is exactly the proof that I‘m talking about in the thread. Again, you do not answer my question. I know and appreciate that you are a master at inequalities but nonetheless, your answer unfortunately does not help me here.
$endgroup$
– Kezer
2 days ago
$begingroup$
@Kezer I just don't know English. :) But I tried to help you.
$endgroup$
– Michael Rozenberg
2 days ago
add a comment |
$begingroup$
By AM-GM
$$left(fraca_1+a_2+...+a_nnright)^n=left(fracfraca_1+a_2+...+a_n-1n-1cdot(n-1)+a_nnright)^ngeq$$
$$geqleft(sqrt[n]left(fraca_1+a_2+...+a_n-1n-1right)^n-1 a_nright)^n=a_nleft(fraca_1+a_2+...+a_n-1n-1right)^n-1.$$
There is also the following proof by Bernoulli.
Let $overlinea=fraca_1+...+a_n-1n-1.$
Thus, $$left(fraca_1+a_2+...+a_nnright)^n=left(fracfraca_1+a_2+...+a_n-1n-1cdot(n-1)+a_nnright)^n=$$
$$=left(frac(n-1)overlinea+a_nnright)^n=overlinea^nleft(1+fraca_nnoverlinea-frac1nright)^ngeqoverlinea^nleft(1+nleft(fraca_nnoverlinea-frac1nright)right)=a_noverlinea^n-1$$
answered Mar 10 at 21:28
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
And again...Brilliant!
$endgroup$
– Dr. Mathva
Mar 10 at 21:31
3
$begingroup$
This does not answer my question and is exactly the proof that I mentioned in the comments.
$endgroup$
– Kezer
Mar 10 at 21:37
1
$begingroup$
The proof by Bernoulli that you edited in is exactly the proof that I‘m talking about in the thread. Again, you do not answer my question. I know and appreciate that you are a master at inequalities but nonetheless, your answer unfortunately does not help me here.
$endgroup$
– Kezer
2 days ago
$begingroup$
@Kezer I just don't know English. :) But I tried to help you.
$endgroup$
– Michael Rozenberg
2 days ago
add a comment |
$begingroup$
By AM-GM
$$left(fraca_1+a_2+...+a_nnright)^n=left(fracfraca_1+a_2+...+a_n-1n-1cdot(n-1)+a_nnright)^ngeq$$
$$geqleft(sqrt[n]left(fraca_1+a_2+...+a_n-1n-1right)^n-1 a_nright)^n=a_nleft(fraca_1+a_2+...+a_n-1n-1right)^n-1.$$
There is also the following proof by Bernoulli.
Let $overlinea=fraca_1+...+a_n-1n-1.$
Thus, $$left(fraca_1+a_2+...+a_nnright)^n=left(fracfraca_1+a_2+...+a_n-1n-1cdot(n-1)+a_nnright)^n=$$
$$=left(frac(n-1)overlinea+a_nnright)^n=overlinea^nleft(1+fraca_nnoverlinea-frac1nright)^ngeqoverlinea^nleft(1+nleft(fraca_nnoverlinea-frac1nright)right)=a_noverlinea^n-1$$
answered Mar 10 at 21:28
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
By AM-GM
$$left(fraca_1+a_2+...+a_nnright)^n=left(fracfraca_1+a_2+...+a_n-1n-1cdot(n-1)+a_nnright)^ngeq$$
$$geqleft(sqrt[n]left(fraca_1+a_2+...+a_n-1n-1right)^n-1 a_nright)^n=a_nleft(fraca_1+a_2+...+a_n-1n-1right)^n-1.$$
There is also the following proof by Bernoulli.
Let $overlinea=fraca_1+...+a_n-1n-1.$
Thus, $$left(fraca_1+a_2+...+a_nnright)^n=left(fracfraca_1+a_2+...+a_n-1n-1cdot(n-1)+a_nnright)^n=$$
$$=left(frac(n-1)overlinea+a_nnright)^n=overlinea^nleft(1+fraca_nnoverlinea-frac1nright)^ngeqoverlinea^nleft(1+nleft(fraca_nnoverlinea-frac1nright)right)=a_noverlinea^n-1$$
answered Mar 10 at 21:28
Michael RozenbergMichael Rozenberg
108k1895200
edited Mar 10 at 21:53
answered Mar 10 at 21:28
Michael RozenbergMichael Rozenberg
108k1895200
answered Mar 10 at 21:28
Michael RozenbergMichael Rozenberg
108k1895200
answered Mar 10 at 21:28
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
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And again...Brilliant!
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– Dr. Mathva
Mar 10 at 21:31
3
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This does not answer my question and is exactly the proof that I mentioned in the comments.
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– Kezer
Mar 10 at 21:37
1
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The proof by Bernoulli that you edited in is exactly the proof that I‘m talking about in the thread. Again, you do not answer my question. I know and appreciate that you are a master at inequalities but nonetheless, your answer unfortunately does not help me here.
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– Kezer
2 days ago
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@Kezer I just don't know English. :) But I tried to help you.
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– Michael Rozenberg
2 days ago
add a comment |
$begingroup$
And again...Brilliant!
$endgroup$
– Dr. Mathva
Mar 10 at 21:31
3
$begingroup$
This does not answer my question and is exactly the proof that I mentioned in the comments.
$endgroup$
– Kezer
Mar 10 at 21:37
1
$begingroup$
The proof by Bernoulli that you edited in is exactly the proof that I‘m talking about in the thread. Again, you do not answer my question. I know and appreciate that you are a master at inequalities but nonetheless, your answer unfortunately does not help me here.
$endgroup$
– Kezer
2 days ago
$begingroup$
@Kezer I just don't know English. :) But I tried to help you.
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
And again...Brilliant!
$endgroup$
– Dr. Mathva
Mar 10 at 21:31
$begingroup$
And again...Brilliant!
$endgroup$
– Dr. Mathva
Mar 10 at 21:31
3
3
$begingroup$
This does not answer my question and is exactly the proof that I mentioned in the comments.
$endgroup$
– Kezer
Mar 10 at 21:37
$begingroup$
This does not answer my question and is exactly the proof that I mentioned in the comments.
$endgroup$
– Kezer
Mar 10 at 21:37
1
1
$begingroup$
The proof by Bernoulli that you edited in is exactly the proof that I‘m talking about in the thread. Again, you do not answer my question. I know and appreciate that you are a master at inequalities but nonetheless, your answer unfortunately does not help me here.
$endgroup$
– Kezer
2 days ago
$begingroup$
The proof by Bernoulli that you edited in is exactly the proof that I‘m talking about in the thread. Again, you do not answer my question. I know and appreciate that you are a master at inequalities but nonetheless, your answer unfortunately does not help me here.
$endgroup$
– Kezer
2 days ago
$begingroup$
@Kezer I just don't know English. :) But I tried to help you.
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
@Kezer I just don't know English. :) But I tried to help you.
$endgroup$
– Michael Rozenberg
2 days ago
add a comment |
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Perhaps also interesting: There is a proof by the weighted AM-GM. Take the $n$-th root. Then, the candidates for the inequality are already written out.
$endgroup$
– Kezer
Mar 10 at 20:52