Double inequality with a certain number of reals
$begingroup$
I've encountered the following problem that I don't know how to solve:
Given positive natural $n$ and positive real $x_1, x_2, ..., x_n$ prove that there exists such positive natural $N$ that
$(1+frac1n)^Nge 2 (x_1+x_2+...+x_{n-1}) + frac{N}nx_nge N^n$.
I don't even know how to start with either of the parts, leave alone both of them simultaneously. I supposed it may be done by the inequality of means but my attempts didn't show anything interesting.
Appreciate your help.
inequality
asked Dec 5 '18 at 19:57
Sorry NorrySorry Norry
342
$endgroup$
add a comment |
$begingroup$
I've encountered the following problem that I don't know how to solve:
Given positive natural $n$ and positive real $x_1, x_2, ..., x_n$ prove that there exists such positive natural $N$ that
$(1+frac1n)^Nge 2 (x_1+x_2+...+x_{n-1}) + frac{N}nx_nge N^n$.
I don't even know how to start with either of the parts, leave alone both of them simultaneously. I supposed it may be done by the inequality of means but my attempts didn't show anything interesting.
Appreciate your help.
inequality
asked Dec 5 '18 at 19:57
Sorry NorrySorry Norry
342
$endgroup$
$begingroup$
@EricTowers x. My bad
$endgroup$
– Sorry Norry
Dec 5 '18 at 20:04
$begingroup$
If $n=2$ and $x_1=x_2=1$ the inequality reduces to $(3/2)^Ngeq N-1geq N^2$, and the second inequality can't be true for any $N$. Do you have a typo somewhere?
$endgroup$
– Joel Moreira
Dec 5 '18 at 20:33
$begingroup$
@JoelMoreira Thank you. I made an error. The correct “developed” fraction in the middle should be reducible to 2. I was too careless. I know it changes a lot. Could you please help me nonetheless?
$endgroup$
– Sorry Norry
Dec 5 '18 at 20:50
$begingroup$
Hm, still when $n=2$ and $x_1=x_2=1/4$ the second inequality becomes $1/2+N/8 geq N^2$ which does not have a solution $N$.
$endgroup$
– Joel Moreira
Dec 5 '18 at 20:59
$begingroup$
More generally, the middle term is linear in $N$ and the right hand side is quadratic when $n=2$, so the second inequality will probably have no solution, unless something changes more dramatically.
$endgroup$
– Joel Moreira
Dec 5 '18 at 21:01
add a comment |
$begingroup$
I've encountered the following problem that I don't know how to solve:
Given positive natural $n$ and positive real $x_1, x_2, ..., x_n$ prove that there exists such positive natural $N$ that
$(1+frac1n)^Nge 2 (x_1+x_2+...+x_{n-1}) + frac{N}nx_nge N^n$.
I don't even know how to start with either of the parts, leave alone both of them simultaneously. I supposed it may be done by the inequality of means but my attempts didn't show anything interesting.
Appreciate your help.
inequality
asked Dec 5 '18 at 19:57
Sorry NorrySorry Norry
342
$endgroup$
I've encountered the following problem that I don't know how to solve:
Given positive natural $n$ and positive real $x_1, x_2, ..., x_n$ prove that there exists such positive natural $N$ that
$(1+frac1n)^Nge 2 (x_1+x_2+...+x_{n-1}) + frac{N}nx_nge N^n$.
I don't even know how to start with either of the parts, leave alone both of them simultaneously. I supposed it may be done by the inequality of means but my attempts didn't show anything interesting.
Appreciate your help.
inequality
inequality
asked Dec 5 '18 at 19:57
Sorry NorrySorry Norry
342
asked Dec 5 '18 at 19:57
Sorry NorrySorry Norry
342
edited Dec 5 '18 at 20:51
Sorry Norry
asked Dec 5 '18 at 19:57
Sorry NorrySorry Norry
342
asked Dec 5 '18 at 19:57
Sorry NorrySorry Norry
342
asked Dec 5 '18 at 19:57
Sorry NorrySorry Norry
342
342
$begingroup$
@EricTowers x. My bad
$endgroup$
– Sorry Norry
Dec 5 '18 at 20:04
$begingroup$
If $n=2$ and $x_1=x_2=1$ the inequality reduces to $(3/2)^Ngeq N-1geq N^2$, and the second inequality can't be true for any $N$. Do you have a typo somewhere?
$endgroup$
– Joel Moreira
Dec 5 '18 at 20:33
$begingroup$
@JoelMoreira Thank you. I made an error. The correct “developed” fraction in the middle should be reducible to 2. I was too careless. I know it changes a lot. Could you please help me nonetheless?
$endgroup$
– Sorry Norry
Dec 5 '18 at 20:50
$begingroup$
Hm, still when $n=2$ and $x_1=x_2=1/4$ the second inequality becomes $1/2+N/8 geq N^2$ which does not have a solution $N$.
$endgroup$
– Joel Moreira
Dec 5 '18 at 20:59
$begingroup$
More generally, the middle term is linear in $N$ and the right hand side is quadratic when $n=2$, so the second inequality will probably have no solution, unless something changes more dramatically.
$endgroup$
– Joel Moreira
Dec 5 '18 at 21:01
add a comment |
$begingroup$
@EricTowers x. My bad
$endgroup$
– Sorry Norry
Dec 5 '18 at 20:04
$begingroup$
If $n=2$ and $x_1=x_2=1$ the inequality reduces to $(3/2)^Ngeq N-1geq N^2$, and the second inequality can't be true for any $N$. Do you have a typo somewhere?
$endgroup$
– Joel Moreira
Dec 5 '18 at 20:33
$begingroup$
@JoelMoreira Thank you. I made an error. The correct “developed” fraction in the middle should be reducible to 2. I was too careless. I know it changes a lot. Could you please help me nonetheless?
$endgroup$
– Sorry Norry
Dec 5 '18 at 20:50
$begingroup$
Hm, still when $n=2$ and $x_1=x_2=1/4$ the second inequality becomes $1/2+N/8 geq N^2$ which does not have a solution $N$.
$endgroup$
– Joel Moreira
Dec 5 '18 at 20:59
$begingroup$
More generally, the middle term is linear in $N$ and the right hand side is quadratic when $n=2$, so the second inequality will probably have no solution, unless something changes more dramatically.
$endgroup$
– Joel Moreira
Dec 5 '18 at 21:01
$begingroup$
@EricTowers x. My bad
$endgroup$
– Sorry Norry
Dec 5 '18 at 20:04
$begingroup$
@EricTowers x. My bad
$endgroup$
– Sorry Norry
Dec 5 '18 at 20:04
$begingroup$
If $n=2$ and $x_1=x_2=1$ the inequality reduces to $(3/2)^Ngeq N-1geq N^2$, and the second inequality can't be true for any $N$. Do you have a typo somewhere?
$endgroup$
– Joel Moreira
Dec 5 '18 at 20:33
$begingroup$
If $n=2$ and $x_1=x_2=1$ the inequality reduces to $(3/2)^Ngeq N-1geq N^2$, and the second inequality can't be true for any $N$. Do you have a typo somewhere?
$endgroup$
– Joel Moreira
Dec 5 '18 at 20:33
$begingroup$
@JoelMoreira Thank you. I made an error. The correct “developed” fraction in the middle should be reducible to 2. I was too careless. I know it changes a lot. Could you please help me nonetheless?
$endgroup$
– Sorry Norry
Dec 5 '18 at 20:50
$begingroup$
@JoelMoreira Thank you. I made an error. The correct “developed” fraction in the middle should be reducible to 2. I was too careless. I know it changes a lot. Could you please help me nonetheless?
$endgroup$
– Sorry Norry
Dec 5 '18 at 20:50
$begingroup$
Hm, still when $n=2$ and $x_1=x_2=1/4$ the second inequality becomes $1/2+N/8 geq N^2$ which does not have a solution $N$.
$endgroup$
– Joel Moreira
Dec 5 '18 at 20:59
$begingroup$
Hm, still when $n=2$ and $x_1=x_2=1/4$ the second inequality becomes $1/2+N/8 geq N^2$ which does not have a solution $N$.
$endgroup$
– Joel Moreira
Dec 5 '18 at 20:59
$begingroup$
More generally, the middle term is linear in $N$ and the right hand side is quadratic when $n=2$, so the second inequality will probably have no solution, unless something changes more dramatically.
$endgroup$
– Joel Moreira
Dec 5 '18 at 21:01
$begingroup$
More generally, the middle term is linear in $N$ and the right hand side is quadratic when $n=2$, so the second inequality will probably have no solution, unless something changes more dramatically.
$endgroup$
– Joel Moreira
Dec 5 '18 at 21:01
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%2f3027569%2fdouble-inequality-with-a-certain-number-of-reals%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%2f3027569%2fdouble-inequality-with-a-certain-number-of-reals%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
$begingroup$
@EricTowers x. My bad
$endgroup$
– Sorry Norry
Dec 5 '18 at 20:04
$begingroup$
If $n=2$ and $x_1=x_2=1$ the inequality reduces to $(3/2)^Ngeq N-1geq N^2$, and the second inequality can't be true for any $N$. Do you have a typo somewhere?
$endgroup$
– Joel Moreira
Dec 5 '18 at 20:33
$begingroup$
@JoelMoreira Thank you. I made an error. The correct “developed” fraction in the middle should be reducible to 2. I was too careless. I know it changes a lot. Could you please help me nonetheless?
$endgroup$
– Sorry Norry
Dec 5 '18 at 20:50
$begingroup$
Hm, still when $n=2$ and $x_1=x_2=1/4$ the second inequality becomes $1/2+N/8 geq N^2$ which does not have a solution $N$.
$endgroup$
– Joel Moreira
Dec 5 '18 at 20:59
$begingroup$
More generally, the middle term is linear in $N$ and the right hand side is quadratic when $n=2$, so the second inequality will probably have no solution, unless something changes more dramatically.
$endgroup$
– Joel Moreira
Dec 5 '18 at 21:01