Does $alpha_j-(Pi alpha_i)^{1/d}=beta_j-(Pi beta_i)^{1/d} $ force $alpha_i=beta_i$?
$begingroup$
Let $d>1$, and let $alpha_1,dots,alpha_d,beta_1,dots,beta_d$ be positive real numbers. Is there an easy proof of the following claim:
If
$alpha_j-(Pi_{i=1}^d alpha_i)^{1/d}=beta_j-(Pi_{i=1}^d beta_i)^{1/d} $ for every $1le jle d$ and not all of these differences are zero, then $alpha_i=beta_i$ for every $i$.
If we allow all of the differences to be zero, then uniqueness fails: set
$alpha_i=a,beta_i=b$ for $a neq b$.
I think I have a proof for this uniqueness, but it is very cumbersome. I am interested in seeing other approaches.
systems-of-equations means a.m.-g.m.-inequality
$endgroup$
add a comment |
$begingroup$
Let $d>1$, and let $alpha_1,dots,alpha_d,beta_1,dots,beta_d$ be positive real numbers. Is there an easy proof of the following claim:
If
$alpha_j-(Pi_{i=1}^d alpha_i)^{1/d}=beta_j-(Pi_{i=1}^d beta_i)^{1/d} $ for every $1le jle d$ and not all of these differences are zero, then $alpha_i=beta_i$ for every $i$.
If we allow all of the differences to be zero, then uniqueness fails: set
$alpha_i=a,beta_i=b$ for $a neq b$.
I think I have a proof for this uniqueness, but it is very cumbersome. I am interested in seeing other approaches.
systems-of-equations means a.m.-g.m.-inequality
$endgroup$
add a comment |
$begingroup$
Let $d>1$, and let $alpha_1,dots,alpha_d,beta_1,dots,beta_d$ be positive real numbers. Is there an easy proof of the following claim:
If
$alpha_j-(Pi_{i=1}^d alpha_i)^{1/d}=beta_j-(Pi_{i=1}^d beta_i)^{1/d} $ for every $1le jle d$ and not all of these differences are zero, then $alpha_i=beta_i$ for every $i$.
If we allow all of the differences to be zero, then uniqueness fails: set
$alpha_i=a,beta_i=b$ for $a neq b$.
I think I have a proof for this uniqueness, but it is very cumbersome. I am interested in seeing other approaches.
systems-of-equations means a.m.-g.m.-inequality
$endgroup$
Let $d>1$, and let $alpha_1,dots,alpha_d,beta_1,dots,beta_d$ be positive real numbers. Is there an easy proof of the following claim:
If
$alpha_j-(Pi_{i=1}^d alpha_i)^{1/d}=beta_j-(Pi_{i=1}^d beta_i)^{1/d} $ for every $1le jle d$ and not all of these differences are zero, then $alpha_i=beta_i$ for every $i$.
If we allow all of the differences to be zero, then uniqueness fails: set
$alpha_i=a,beta_i=b$ for $a neq b$.
I think I have a proof for this uniqueness, but it is very cumbersome. I am interested in seeing other approaches.
systems-of-equations means a.m.-g.m.-inequality
systems-of-equations means a.m.-g.m.-inequality
asked Dec 11 '18 at 10:08
Asaf ShacharAsaf Shachar
5,64631141
5,64631141
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1 Answer
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$begingroup$
Assume that that condition holds. Then $beta_i=c+alpha_i$ for all $i,$ where $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$ If $c=0$ we're done, so assume not. By swapping $alpha$ and $beta$ we can assume $c> 0$ without loss of generality.
The $d$-function $L^d$ Hölder inequality applied to the functions $f_i$ defined by $f_i(0)=c^{1/d}$ and $f_i(1)=alpha_i^{1/d},$ on the domain ${0,1}$ with counting measure, gives
$$c+prod_{i=1}^d alpha_i^{1/d}leq prod_{i=1}^d |f_i|_d=prod_{i=1}^d (c+alpha_i)^{1/d}=prod_{i=1}^d beta_i^{1/d}$$
with strict inequality unless all the $alpha_i$ are the same. But strict inequality would contradict $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$
$endgroup$
$begingroup$
Wow! How on earth did you think about this use of Holder's inequality? This seems like a very non-trivial approach...Thank you.
$endgroup$
– Asaf Shachar
Dec 11 '18 at 15:02
$begingroup$
Hölder's inequality for finite sums is a standard tool in "problem solving"/"contest math" style inequalities. Though I had happened to have just read another question where its use was quite surprising to me: math.stackexchange.com/questions/3031422/…
$endgroup$
– Dap
Dec 11 '18 at 18:55
$begingroup$
Thank you. I still find this use of Hölder's inequality somewhat miraculous... so I asked another question about it here: math.stackexchange.com/questions/3036749/…. Thank you again for this very nice solution.
$endgroup$
– Asaf Shachar
Dec 12 '18 at 14:29
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Assume that that condition holds. Then $beta_i=c+alpha_i$ for all $i,$ where $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$ If $c=0$ we're done, so assume not. By swapping $alpha$ and $beta$ we can assume $c> 0$ without loss of generality.
The $d$-function $L^d$ Hölder inequality applied to the functions $f_i$ defined by $f_i(0)=c^{1/d}$ and $f_i(1)=alpha_i^{1/d},$ on the domain ${0,1}$ with counting measure, gives
$$c+prod_{i=1}^d alpha_i^{1/d}leq prod_{i=1}^d |f_i|_d=prod_{i=1}^d (c+alpha_i)^{1/d}=prod_{i=1}^d beta_i^{1/d}$$
with strict inequality unless all the $alpha_i$ are the same. But strict inequality would contradict $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$
$endgroup$
$begingroup$
Wow! How on earth did you think about this use of Holder's inequality? This seems like a very non-trivial approach...Thank you.
$endgroup$
– Asaf Shachar
Dec 11 '18 at 15:02
$begingroup$
Hölder's inequality for finite sums is a standard tool in "problem solving"/"contest math" style inequalities. Though I had happened to have just read another question where its use was quite surprising to me: math.stackexchange.com/questions/3031422/…
$endgroup$
– Dap
Dec 11 '18 at 18:55
$begingroup$
Thank you. I still find this use of Hölder's inequality somewhat miraculous... so I asked another question about it here: math.stackexchange.com/questions/3036749/…. Thank you again for this very nice solution.
$endgroup$
– Asaf Shachar
Dec 12 '18 at 14:29
add a comment |
$begingroup$
Assume that that condition holds. Then $beta_i=c+alpha_i$ for all $i,$ where $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$ If $c=0$ we're done, so assume not. By swapping $alpha$ and $beta$ we can assume $c> 0$ without loss of generality.
The $d$-function $L^d$ Hölder inequality applied to the functions $f_i$ defined by $f_i(0)=c^{1/d}$ and $f_i(1)=alpha_i^{1/d},$ on the domain ${0,1}$ with counting measure, gives
$$c+prod_{i=1}^d alpha_i^{1/d}leq prod_{i=1}^d |f_i|_d=prod_{i=1}^d (c+alpha_i)^{1/d}=prod_{i=1}^d beta_i^{1/d}$$
with strict inequality unless all the $alpha_i$ are the same. But strict inequality would contradict $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$
$endgroup$
$begingroup$
Wow! How on earth did you think about this use of Holder's inequality? This seems like a very non-trivial approach...Thank you.
$endgroup$
– Asaf Shachar
Dec 11 '18 at 15:02
$begingroup$
Hölder's inequality for finite sums is a standard tool in "problem solving"/"contest math" style inequalities. Though I had happened to have just read another question where its use was quite surprising to me: math.stackexchange.com/questions/3031422/…
$endgroup$
– Dap
Dec 11 '18 at 18:55
$begingroup$
Thank you. I still find this use of Hölder's inequality somewhat miraculous... so I asked another question about it here: math.stackexchange.com/questions/3036749/…. Thank you again for this very nice solution.
$endgroup$
– Asaf Shachar
Dec 12 '18 at 14:29
add a comment |
$begingroup$
Assume that that condition holds. Then $beta_i=c+alpha_i$ for all $i,$ where $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$ If $c=0$ we're done, so assume not. By swapping $alpha$ and $beta$ we can assume $c> 0$ without loss of generality.
The $d$-function $L^d$ Hölder inequality applied to the functions $f_i$ defined by $f_i(0)=c^{1/d}$ and $f_i(1)=alpha_i^{1/d},$ on the domain ${0,1}$ with counting measure, gives
$$c+prod_{i=1}^d alpha_i^{1/d}leq prod_{i=1}^d |f_i|_d=prod_{i=1}^d (c+alpha_i)^{1/d}=prod_{i=1}^d beta_i^{1/d}$$
with strict inequality unless all the $alpha_i$ are the same. But strict inequality would contradict $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$
$endgroup$
Assume that that condition holds. Then $beta_i=c+alpha_i$ for all $i,$ where $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$ If $c=0$ we're done, so assume not. By swapping $alpha$ and $beta$ we can assume $c> 0$ without loss of generality.
The $d$-function $L^d$ Hölder inequality applied to the functions $f_i$ defined by $f_i(0)=c^{1/d}$ and $f_i(1)=alpha_i^{1/d},$ on the domain ${0,1}$ with counting measure, gives
$$c+prod_{i=1}^d alpha_i^{1/d}leq prod_{i=1}^d |f_i|_d=prod_{i=1}^d (c+alpha_i)^{1/d}=prod_{i=1}^d beta_i^{1/d}$$
with strict inequality unless all the $alpha_i$ are the same. But strict inequality would contradict $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$
edited Dec 11 '18 at 12:40
answered Dec 11 '18 at 10:51
DapDap
17.3k840
17.3k840
$begingroup$
Wow! How on earth did you think about this use of Holder's inequality? This seems like a very non-trivial approach...Thank you.
$endgroup$
– Asaf Shachar
Dec 11 '18 at 15:02
$begingroup$
Hölder's inequality for finite sums is a standard tool in "problem solving"/"contest math" style inequalities. Though I had happened to have just read another question where its use was quite surprising to me: math.stackexchange.com/questions/3031422/…
$endgroup$
– Dap
Dec 11 '18 at 18:55
$begingroup$
Thank you. I still find this use of Hölder's inequality somewhat miraculous... so I asked another question about it here: math.stackexchange.com/questions/3036749/…. Thank you again for this very nice solution.
$endgroup$
– Asaf Shachar
Dec 12 '18 at 14:29
add a comment |
$begingroup$
Wow! How on earth did you think about this use of Holder's inequality? This seems like a very non-trivial approach...Thank you.
$endgroup$
– Asaf Shachar
Dec 11 '18 at 15:02
$begingroup$
Hölder's inequality for finite sums is a standard tool in "problem solving"/"contest math" style inequalities. Though I had happened to have just read another question where its use was quite surprising to me: math.stackexchange.com/questions/3031422/…
$endgroup$
– Dap
Dec 11 '18 at 18:55
$begingroup$
Thank you. I still find this use of Hölder's inequality somewhat miraculous... so I asked another question about it here: math.stackexchange.com/questions/3036749/…. Thank you again for this very nice solution.
$endgroup$
– Asaf Shachar
Dec 12 '18 at 14:29
$begingroup$
Wow! How on earth did you think about this use of Holder's inequality? This seems like a very non-trivial approach...Thank you.
$endgroup$
– Asaf Shachar
Dec 11 '18 at 15:02
$begingroup$
Wow! How on earth did you think about this use of Holder's inequality? This seems like a very non-trivial approach...Thank you.
$endgroup$
– Asaf Shachar
Dec 11 '18 at 15:02
$begingroup$
Hölder's inequality for finite sums is a standard tool in "problem solving"/"contest math" style inequalities. Though I had happened to have just read another question where its use was quite surprising to me: math.stackexchange.com/questions/3031422/…
$endgroup$
– Dap
Dec 11 '18 at 18:55
$begingroup$
Hölder's inequality for finite sums is a standard tool in "problem solving"/"contest math" style inequalities. Though I had happened to have just read another question where its use was quite surprising to me: math.stackexchange.com/questions/3031422/…
$endgroup$
– Dap
Dec 11 '18 at 18:55
$begingroup$
Thank you. I still find this use of Hölder's inequality somewhat miraculous... so I asked another question about it here: math.stackexchange.com/questions/3036749/…. Thank you again for this very nice solution.
$endgroup$
– Asaf Shachar
Dec 12 '18 at 14:29
$begingroup$
Thank you. I still find this use of Hölder's inequality somewhat miraculous... so I asked another question about it here: math.stackexchange.com/questions/3036749/…. Thank you again for this very nice solution.
$endgroup$
– Asaf Shachar
Dec 12 '18 at 14:29
add a comment |
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