$N-1$ equal value principle
$begingroup$
Let us start with an example, consider the following sum: $$H_n=sum_{i=1}^{n}frac{1}{n-1+x_i}$$ With $x_1x_2...x_n=1$, We're willing to show that $H_n le 1$.
To do this we start with the following lemma:
Lemma:
If a differentiable function has a single inflexion
point and is evaluated at n arbitrary reals with a fixed sum, any minimum or maximum must occur where
some $n − 1$ variables are equal.
Proof [From mildorf introduction to inequalities]:
Consider $f(y)=frac{1}{k+e^y}$ for a non-negative constant $k$. We have $f''(y)=frac{e^y(e^y-k)}{(k+e^y)^3}$, so $f''(y)ge 0 iff e^yge k$, thus $f(y)$ has exactly one inflexion point at $y=ln{k}$ and $f$ is convex in the interval $(ln k,infty) $. Consider now $y_i=ln{(x_i)}$ therefore $y_1+y_2+...y_n=0$, and if we take $k=n-1$ we have:$$H_n=sum_{i=1}^{n}f(y_i)$$ The inflection point will be at $k_0=ln(k)=ln(n-1)$ Suppose $y_1 ge y_2 ge...ge y_m ge k_0 ge y_{m+1}... ge y_n $ for some positive $m$. Then by Karamata's Inequality (also called majorization inequality):$$f(y_1)+f(y_2)+...f(y_m) le (m-1)f(k_0)+f(y_1+...+y_m-(m-1)k_0) :::[1] $$
Also by Karamata: $$(m-1)f(k_0)+f(y_{m+1})+...+f(y_n) le (n-1)f(frac{(m-1)k_0+y_{m+1}+...+y_n}{n-1}) ::: [2]$$
"otherwise, all of the $y_i$ are less than $k_0$ In that case we may directly apply Majorization
to equate $n-1$ of the $y_i$ whilst increasing the sum in question. Hence, the lemma is
valid"
With this lemma we can prove our inequality quite easly, so I'll skip on that. My question is, in the proof I understood everything up to the very end, I've put quotation marks on the part that confused me, I really don't see his arguement or what he is trying to say in those few lines and I'm kinda lost on why does this prove our initial statement, any help is appreciated!
inequality karamata-inequality
edited Dec 15 '18 at 9:19
Michael Rozenberg
102k1791195
asked Dec 14 '18 at 17:10
Spasoje DurovicSpasoje Durovic
36010
$endgroup$
|
show 3 more comments
$begingroup$
Let us start with an example, consider the following sum: $$H_n=sum_{i=1}^{n}frac{1}{n-1+x_i}$$ With $x_1x_2...x_n=1$, We're willing to show that $H_n le 1$.
To do this we start with the following lemma:
Lemma:
If a differentiable function has a single inflexion
point and is evaluated at n arbitrary reals with a fixed sum, any minimum or maximum must occur where
some $n − 1$ variables are equal.
Proof [From mildorf introduction to inequalities]:
Consider $f(y)=frac{1}{k+e^y}$ for a non-negative constant $k$. We have $f''(y)=frac{e^y(e^y-k)}{(k+e^y)^3}$, so $f''(y)ge 0 iff e^yge k$, thus $f(y)$ has exactly one inflexion point at $y=ln{k}$ and $f$ is convex in the interval $(ln k,infty) $. Consider now $y_i=ln{(x_i)}$ therefore $y_1+y_2+...y_n=0$, and if we take $k=n-1$ we have:$$H_n=sum_{i=1}^{n}f(y_i)$$ The inflection point will be at $k_0=ln(k)=ln(n-1)$ Suppose $y_1 ge y_2 ge...ge y_m ge k_0 ge y_{m+1}... ge y_n $ for some positive $m$. Then by Karamata's Inequality (also called majorization inequality):$$f(y_1)+f(y_2)+...f(y_m) le (m-1)f(k_0)+f(y_1+...+y_m-(m-1)k_0) :::[1] $$
Also by Karamata: $$(m-1)f(k_0)+f(y_{m+1})+...+f(y_n) le (n-1)f(frac{(m-1)k_0+y_{m+1}+...+y_n}{n-1}) ::: [2]$$
"otherwise, all of the $y_i$ are less than $k_0$ In that case we may directly apply Majorization
to equate $n-1$ of the $y_i$ whilst increasing the sum in question. Hence, the lemma is
valid"
With this lemma we can prove our inequality quite easly, so I'll skip on that. My question is, in the proof I understood everything up to the very end, I've put quotation marks on the part that confused me, I really don't see his arguement or what he is trying to say in those few lines and I'm kinda lost on why does this prove our initial statement, any help is appreciated!
inequality karamata-inequality
edited Dec 15 '18 at 9:19
Michael Rozenberg
102k1791195
asked Dec 14 '18 at 17:10
Spasoje DurovicSpasoje Durovic
36010
$endgroup$
1
$begingroup$
Read that as "Suppose ____ for some positive $m$. Then _____blah blah____. Otherwise ..." does that help you see the argument? If there is no such positive $m$, it is a simple case of every variable being in the concave domain, so maximisation is a trivial case of Karamata (or Jensen).
$endgroup$
– Macavity
Dec 14 '18 at 18:36
1
$begingroup$
Yep, that looks right. Now you may want to try your example and see.
$endgroup$
– Macavity
Dec 14 '18 at 18:56
1
$begingroup$
@Spasoje Durovic There is a very nice proof of this inequality by AM-GM. By the way, this inequality is true also by the Vasc's LCF Theorem.
$endgroup$
– Michael Rozenberg
Dec 14 '18 at 18:59
1
$begingroup$
@Hanno I edited it as you suggested :) Personally I didn't find the proof he was talking about but I did find the "Vasc's LCF Theorem" he mentioned, if you're interessed just google "generalization of Jensens Inequality Gazeta Matematica" It should be the second result, it seems quite interessting
$endgroup$
– Spasoje Durovic
Dec 15 '18 at 8:41
2
$begingroup$
@Hanno Here you go... artofproblemsolving.com/community/c6h224946p1249268
$endgroup$
– Macavity
Dec 15 '18 at 10:06
|
show 3 more comments
$begingroup$
Let us start with an example, consider the following sum: $$H_n=sum_{i=1}^{n}frac{1}{n-1+x_i}$$ With $x_1x_2...x_n=1$, We're willing to show that $H_n le 1$.
To do this we start with the following lemma:
Lemma:
If a differentiable function has a single inflexion
point and is evaluated at n arbitrary reals with a fixed sum, any minimum or maximum must occur where
some $n − 1$ variables are equal.
Proof [From mildorf introduction to inequalities]:
Consider $f(y)=frac{1}{k+e^y}$ for a non-negative constant $k$. We have $f''(y)=frac{e^y(e^y-k)}{(k+e^y)^3}$, so $f''(y)ge 0 iff e^yge k$, thus $f(y)$ has exactly one inflexion point at $y=ln{k}$ and $f$ is convex in the interval $(ln k,infty) $. Consider now $y_i=ln{(x_i)}$ therefore $y_1+y_2+...y_n=0$, and if we take $k=n-1$ we have:$$H_n=sum_{i=1}^{n}f(y_i)$$ The inflection point will be at $k_0=ln(k)=ln(n-1)$ Suppose $y_1 ge y_2 ge...ge y_m ge k_0 ge y_{m+1}... ge y_n $ for some positive $m$. Then by Karamata's Inequality (also called majorization inequality):$$f(y_1)+f(y_2)+...f(y_m) le (m-1)f(k_0)+f(y_1+...+y_m-(m-1)k_0) :::[1] $$
Also by Karamata: $$(m-1)f(k_0)+f(y_{m+1})+...+f(y_n) le (n-1)f(frac{(m-1)k_0+y_{m+1}+...+y_n}{n-1}) ::: [2]$$
"otherwise, all of the $y_i$ are less than $k_0$ In that case we may directly apply Majorization
to equate $n-1$ of the $y_i$ whilst increasing the sum in question. Hence, the lemma is
valid"
With this lemma we can prove our inequality quite easly, so I'll skip on that. My question is, in the proof I understood everything up to the very end, I've put quotation marks on the part that confused me, I really don't see his arguement or what he is trying to say in those few lines and I'm kinda lost on why does this prove our initial statement, any help is appreciated!
inequality karamata-inequality
edited Dec 15 '18 at 9:19
Michael Rozenberg
102k1791195
asked Dec 14 '18 at 17:10
Spasoje DurovicSpasoje Durovic
36010
$endgroup$
Let us start with an example, consider the following sum: $$H_n=sum_{i=1}^{n}frac{1}{n-1+x_i}$$ With $x_1x_2...x_n=1$, We're willing to show that $H_n le 1$.
To do this we start with the following lemma:
Lemma:
If a differentiable function has a single inflexion
point and is evaluated at n arbitrary reals with a fixed sum, any minimum or maximum must occur where
some $n − 1$ variables are equal.
Proof [From mildorf introduction to inequalities]:
Consider $f(y)=frac{1}{k+e^y}$ for a non-negative constant $k$. We have $f''(y)=frac{e^y(e^y-k)}{(k+e^y)^3}$, so $f''(y)ge 0 iff e^yge k$, thus $f(y)$ has exactly one inflexion point at $y=ln{k}$ and $f$ is convex in the interval $(ln k,infty) $. Consider now $y_i=ln{(x_i)}$ therefore $y_1+y_2+...y_n=0$, and if we take $k=n-1$ we have:$$H_n=sum_{i=1}^{n}f(y_i)$$ The inflection point will be at $k_0=ln(k)=ln(n-1)$ Suppose $y_1 ge y_2 ge...ge y_m ge k_0 ge y_{m+1}... ge y_n $ for some positive $m$. Then by Karamata's Inequality (also called majorization inequality):$$f(y_1)+f(y_2)+...f(y_m) le (m-1)f(k_0)+f(y_1+...+y_m-(m-1)k_0) :::[1] $$
Also by Karamata: $$(m-1)f(k_0)+f(y_{m+1})+...+f(y_n) le (n-1)f(frac{(m-1)k_0+y_{m+1}+...+y_n}{n-1}) ::: [2]$$
"otherwise, all of the $y_i$ are less than $k_0$ In that case we may directly apply Majorization
to equate $n-1$ of the $y_i$ whilst increasing the sum in question. Hence, the lemma is
valid"
With this lemma we can prove our inequality quite easly, so I'll skip on that. My question is, in the proof I understood everything up to the very end, I've put quotation marks on the part that confused me, I really don't see his arguement or what he is trying to say in those few lines and I'm kinda lost on why does this prove our initial statement, any help is appreciated!
inequality karamata-inequality
inequality karamata-inequality
edited Dec 15 '18 at 9:19
Michael Rozenberg
102k1791195
asked Dec 14 '18 at 17:10
Spasoje DurovicSpasoje Durovic
36010
edited Dec 15 '18 at 9:19
Michael Rozenberg
102k1791195
asked Dec 14 '18 at 17:10
Spasoje DurovicSpasoje Durovic
36010
edited Dec 15 '18 at 9:19
Michael Rozenberg
102k1791195
edited Dec 15 '18 at 9:19
Michael Rozenberg
102k1791195
edited Dec 15 '18 at 9:19
Michael Rozenberg
102k1791195
102k1791195
asked Dec 14 '18 at 17:10
Spasoje DurovicSpasoje Durovic
36010
asked Dec 14 '18 at 17:10
Spasoje DurovicSpasoje Durovic
36010
asked Dec 14 '18 at 17:10
Spasoje DurovicSpasoje Durovic
36010
36010
1
$begingroup$
Read that as "Suppose ____ for some positive $m$. Then _____blah blah____. Otherwise ..." does that help you see the argument? If there is no such positive $m$, it is a simple case of every variable being in the concave domain, so maximisation is a trivial case of Karamata (or Jensen).
$endgroup$
– Macavity
Dec 14 '18 at 18:36
1
$begingroup$
Yep, that looks right. Now you may want to try your example and see.
$endgroup$
– Macavity
Dec 14 '18 at 18:56
1
$begingroup$
@Spasoje Durovic There is a very nice proof of this inequality by AM-GM. By the way, this inequality is true also by the Vasc's LCF Theorem.
$endgroup$
– Michael Rozenberg
Dec 14 '18 at 18:59
1
$begingroup$
@Hanno I edited it as you suggested :) Personally I didn't find the proof he was talking about but I did find the "Vasc's LCF Theorem" he mentioned, if you're interessed just google "generalization of Jensens Inequality Gazeta Matematica" It should be the second result, it seems quite interessting
$endgroup$
– Spasoje Durovic
Dec 15 '18 at 8:41
2
$begingroup$
@Hanno Here you go... artofproblemsolving.com/community/c6h224946p1249268
$endgroup$
– Macavity
Dec 15 '18 at 10:06
|
show 3 more comments
1
$begingroup$
Read that as "Suppose ____ for some positive $m$. Then _____blah blah____. Otherwise ..." does that help you see the argument? If there is no such positive $m$, it is a simple case of every variable being in the concave domain, so maximisation is a trivial case of Karamata (or Jensen).
$endgroup$
– Macavity
Dec 14 '18 at 18:36
1
$begingroup$
Yep, that looks right. Now you may want to try your example and see.
$endgroup$
– Macavity
Dec 14 '18 at 18:56
1
$begingroup$
@Spasoje Durovic There is a very nice proof of this inequality by AM-GM. By the way, this inequality is true also by the Vasc's LCF Theorem.
$endgroup$
– Michael Rozenberg
Dec 14 '18 at 18:59
1
$begingroup$
@Hanno I edited it as you suggested :) Personally I didn't find the proof he was talking about but I did find the "Vasc's LCF Theorem" he mentioned, if you're interessed just google "generalization of Jensens Inequality Gazeta Matematica" It should be the second result, it seems quite interessting
$endgroup$
– Spasoje Durovic
Dec 15 '18 at 8:41
2
$begingroup$
@Hanno Here you go... artofproblemsolving.com/community/c6h224946p1249268
$endgroup$
– Macavity
Dec 15 '18 at 10:06
1
1
$begingroup$
Read that as "Suppose ____ for some positive $m$. Then _____blah blah____. Otherwise ..." does that help you see the argument? If there is no such positive $m$, it is a simple case of every variable being in the concave domain, so maximisation is a trivial case of Karamata (or Jensen).
$endgroup$
– Macavity
Dec 14 '18 at 18:36
$begingroup$
Read that as "Suppose ____ for some positive $m$. Then _____blah blah____. Otherwise ..." does that help you see the argument? If there is no such positive $m$, it is a simple case of every variable being in the concave domain, so maximisation is a trivial case of Karamata (or Jensen).
$endgroup$
– Macavity
Dec 14 '18 at 18:36
1
1
$begingroup$
Yep, that looks right. Now you may want to try your example and see.
$endgroup$
– Macavity
Dec 14 '18 at 18:56
$begingroup$
Yep, that looks right. Now you may want to try your example and see.
$endgroup$
– Macavity
Dec 14 '18 at 18:56
1
1
$begingroup$
@Spasoje Durovic There is a very nice proof of this inequality by AM-GM. By the way, this inequality is true also by the Vasc's LCF Theorem.
$endgroup$
– Michael Rozenberg
Dec 14 '18 at 18:59
$begingroup$
@Spasoje Durovic There is a very nice proof of this inequality by AM-GM. By the way, this inequality is true also by the Vasc's LCF Theorem.
$endgroup$
– Michael Rozenberg
Dec 14 '18 at 18:59
1
1
$begingroup$
@Hanno I edited it as you suggested :) Personally I didn't find the proof he was talking about but I did find the "Vasc's LCF Theorem" he mentioned, if you're interessed just google "generalization of Jensens Inequality Gazeta Matematica" It should be the second result, it seems quite interessting
$endgroup$
– Spasoje Durovic
Dec 15 '18 at 8:41
$begingroup$
@Hanno I edited it as you suggested :) Personally I didn't find the proof he was talking about but I did find the "Vasc's LCF Theorem" he mentioned, if you're interessed just google "generalization of Jensens Inequality Gazeta Matematica" It should be the second result, it seems quite interessting
$endgroup$
– Spasoje Durovic
Dec 15 '18 at 8:41
2
2
$begingroup$
@Hanno Here you go... artofproblemsolving.com/community/c6h224946p1249268
$endgroup$
– Macavity
Dec 15 '18 at 10:06
$begingroup$
@Hanno Here you go... artofproblemsolving.com/community/c6h224946p1249268
$endgroup$
– Macavity
Dec 15 '18 at 10:06
|
show 3 more comments
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039653%2fn-1-equal-value-principle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039653%2fn-1-equal-value-principle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Read that as "Suppose ____ for some positive $m$. Then _____blah blah____. Otherwise ..." does that help you see the argument? If there is no such positive $m$, it is a simple case of every variable being in the concave domain, so maximisation is a trivial case of Karamata (or Jensen).
$endgroup$
– Macavity
Dec 14 '18 at 18:36
1
$begingroup$
Yep, that looks right. Now you may want to try your example and see.
$endgroup$
– Macavity
Dec 14 '18 at 18:56
1
$begingroup$
@Spasoje Durovic There is a very nice proof of this inequality by AM-GM. By the way, this inequality is true also by the Vasc's LCF Theorem.
$endgroup$
– Michael Rozenberg
Dec 14 '18 at 18:59
1
$begingroup$
@Hanno I edited it as you suggested :) Personally I didn't find the proof he was talking about but I did find the "Vasc's LCF Theorem" he mentioned, if you're interessed just google "generalization of Jensens Inequality Gazeta Matematica" It should be the second result, it seems quite interessting
$endgroup$
– Spasoje Durovic
Dec 15 '18 at 8:41
2
$begingroup$
@Hanno Here you go... artofproblemsolving.com/community/c6h224946p1249268
$endgroup$
– Macavity
Dec 15 '18 at 10:06