Mixing
Stronger than Karamata
$begingroup$
Hello I have this to propose :
Let $x,y,a,b>0$ be real positive numbers and $f(x)$ be a convex increasing function with $f(0) > 0$ such that $f(x)geq f(a) geq f(b)geq f(y)geq x geq ageq b geq y $ and $x+y geq a+b$ then we have :
$$f(x)+f(y)-frac{1}{2}f(frac{x+y}{2sqrt{xy}})geq f(a)+f(b)-frac{1}{2}f(frac{a+b}{2sqrt{ab}}) $$
I have tried a lot of majorization but without success , and the problem seems hard .And I don't now if $frac{1}{2}$ is the best constant in this problem .
So can someone prove this ?
Thanks in advance .
real-analysis inequality convex-analysis
$endgroup$
add a comment |
$begingroup$
Hello I have this to propose :
Let $x,y,a,b>0$ be real positive numbers and $f(x)$ be a convex increasing function with $f(0) > 0$ such that $f(x)geq f(a) geq f(b)geq f(y)geq x geq ageq b geq y $ and $x+y geq a+b$ then we have :
$$f(x)+f(y)-frac{1}{2}f(frac{x+y}{2sqrt{xy}})geq f(a)+f(b)-frac{1}{2}f(frac{a+b}{2sqrt{ab}}) $$
I have tried a lot of majorization but without success , and the problem seems hard .And I don't now if $frac{1}{2}$ is the best constant in this problem .
So can someone prove this ?
Thanks in advance .
real-analysis inequality convex-analysis
$endgroup$
2
$begingroup$
Since you love to change the given in your problems, I'll say only: Your statement is wrong.
$endgroup$
– Michael Rozenberg
Jan 12 at 15:03
$begingroup$
I'm voting to close this question as off-topic because it asks to prove a falsehood.
$endgroup$
– Macavity
Jan 12 at 16:56
$begingroup$
@Macavity Why is it wrong? It's not immediately obvious to me
$endgroup$
– user574848
Jan 12 at 22:51
2
$begingroup$
@user574848 The real question is, why is it right? The question shows no effort at establishing why it could be right - it is just a blind conjecture thrown for others to work on. In fact if you work on it, it is not hard to come up with counter examples - I am not inclined to help the questioner this time, did it for the last question and the OP just changed the constants in the question to other random values. Just reminds me of trolling.
$endgroup$
– Macavity
Jan 13 at 13:46
add a comment |
$begingroup$
Hello I have this to propose :
Let $x,y,a,b>0$ be real positive numbers and $f(x)$ be a convex increasing function with $f(0) > 0$ such that $f(x)geq f(a) geq f(b)geq f(y)geq x geq ageq b geq y $ and $x+y geq a+b$ then we have :
$$f(x)+f(y)-frac{1}{2}f(frac{x+y}{2sqrt{xy}})geq f(a)+f(b)-frac{1}{2}f(frac{a+b}{2sqrt{ab}}) $$
I have tried a lot of majorization but without success , and the problem seems hard .And I don't now if $frac{1}{2}$ is the best constant in this problem .
So can someone prove this ?
Thanks in advance .
real-analysis inequality convex-analysis
$endgroup$
Hello I have this to propose :
Let $x,y,a,b>0$ be real positive numbers and $f(x)$ be a convex increasing function with $f(0) > 0$ such that $f(x)geq f(a) geq f(b)geq f(y)geq x geq ageq b geq y $ and $x+y geq a+b$ then we have :
$$f(x)+f(y)-frac{1}{2}f(frac{x+y}{2sqrt{xy}})geq f(a)+f(b)-frac{1}{2}f(frac{a+b}{2sqrt{ab}}) $$
I have tried a lot of majorization but without success , and the problem seems hard .And I don't now if $frac{1}{2}$ is the best constant in this problem .
So can someone prove this ?
Thanks in advance .
real-analysis inequality convex-analysis
real-analysis inequality convex-analysis
asked Jan 12 at 13:47
user448747
2
$begingroup$
Since you love to change the given in your problems, I'll say only: Your statement is wrong.
$endgroup$
– Michael Rozenberg
Jan 12 at 15:03
$begingroup$
I'm voting to close this question as off-topic because it asks to prove a falsehood.
$endgroup$
– Macavity
Jan 12 at 16:56
$begingroup$
@Macavity Why is it wrong? It's not immediately obvious to me
$endgroup$
– user574848
Jan 12 at 22:51
2
$begingroup$
@user574848 The real question is, why is it right? The question shows no effort at establishing why it could be right - it is just a blind conjecture thrown for others to work on. In fact if you work on it, it is not hard to come up with counter examples - I am not inclined to help the questioner this time, did it for the last question and the OP just changed the constants in the question to other random values. Just reminds me of trolling.
$endgroup$
– Macavity
Jan 13 at 13:46
add a comment |
2
$begingroup$
Since you love to change the given in your problems, I'll say only: Your statement is wrong.
$endgroup$
– Michael Rozenberg
Jan 12 at 15:03
$begingroup$
I'm voting to close this question as off-topic because it asks to prove a falsehood.
$endgroup$
– Macavity
Jan 12 at 16:56
$begingroup$
@Macavity Why is it wrong? It's not immediately obvious to me
$endgroup$
– user574848
Jan 12 at 22:51
2
$begingroup$
@user574848 The real question is, why is it right? The question shows no effort at establishing why it could be right - it is just a blind conjecture thrown for others to work on. In fact if you work on it, it is not hard to come up with counter examples - I am not inclined to help the questioner this time, did it for the last question and the OP just changed the constants in the question to other random values. Just reminds me of trolling.
$endgroup$
– Macavity
Jan 13 at 13:46
2
2
$begingroup$
Since you love to change the given in your problems, I'll say only: Your statement is wrong.
$endgroup$
– Michael Rozenberg
Jan 12 at 15:03
$begingroup$
Since you love to change the given in your problems, I'll say only: Your statement is wrong.
$endgroup$
– Michael Rozenberg
Jan 12 at 15:03
$begingroup$
I'm voting to close this question as off-topic because it asks to prove a falsehood.
$endgroup$
– Macavity
Jan 12 at 16:56
$begingroup$
I'm voting to close this question as off-topic because it asks to prove a falsehood.
$endgroup$
– Macavity
Jan 12 at 16:56
$begingroup$
@Macavity Why is it wrong? It's not immediately obvious to me
$endgroup$
– user574848
Jan 12 at 22:51
$begingroup$
@Macavity Why is it wrong? It's not immediately obvious to me
$endgroup$
– user574848
Jan 12 at 22:51
2
2
$begingroup$
@user574848 The real question is, why is it right? The question shows no effort at establishing why it could be right - it is just a blind conjecture thrown for others to work on. In fact if you work on it, it is not hard to come up with counter examples - I am not inclined to help the questioner this time, did it for the last question and the OP just changed the constants in the question to other random values. Just reminds me of trolling.
$endgroup$
– Macavity
Jan 13 at 13:46
$begingroup$
@user574848 The real question is, why is it right? The question shows no effort at establishing why it could be right - it is just a blind conjecture thrown for others to work on. In fact if you work on it, it is not hard to come up with counter examples - I am not inclined to help the questioner this time, did it for the last question and the OP just changed the constants in the question to other random values. Just reminds me of trolling.
$endgroup$
– Macavity
Jan 13 at 13:46
add a comment |
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%2f3070912%2fstronger-than-karamata%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%2f3070912%2fstronger-than-karamata%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
2
$begingroup$
Since you love to change the given in your problems, I'll say only: Your statement is wrong.
$endgroup$
– Michael Rozenberg
Jan 12 at 15:03
$begingroup$
I'm voting to close this question as off-topic because it asks to prove a falsehood.
$endgroup$
– Macavity
Jan 12 at 16:56
$begingroup$
@Macavity Why is it wrong? It's not immediately obvious to me
$endgroup$
– user574848
Jan 12 at 22:51
2
$begingroup$
@user574848 The real question is, why is it right? The question shows no effort at establishing why it could be right - it is just a blind conjecture thrown for others to work on. In fact if you work on it, it is not hard to come up with counter examples - I am not inclined to help the questioner this time, did it for the last question and the OP just changed the constants in the question to other random values. Just reminds me of trolling.
$endgroup$
– Macavity
Jan 13 at 13:46