Mixing
chebyshev's inequality - Question
I had a question in my exam and they asked to prove that
prove that:
$$3(1+a^2+a^4)geq(1+a+a^2)^2$$ for all $ainmathbb R$.
Now , I solved it , but the problem is that in the answer they wrote this:
using Chebyshev inequality:
$$(1+a+a^2)^2=(1·1+a·1+a^2·1)^2≤(1+a^2+a^4)·(1+1+1)=3(1+a^2+a^4).$$
And so I tried searching the web for this inequality but all it found was the Chebyshev's inequality for probabillity.
can someone please send me link regarding this inequality or just write it here?
Thank you.
algebra-precalculus inequality symmetric-polynomials rearrangement-inequality
edited Jan 1 at 1:24
Michael Rozenberg
97.5k1589188
asked Dec 31 '18 at 23:59
orel Zemachorel Zemach
232
add a comment |
I had a question in my exam and they asked to prove that
prove that:
$$3(1+a^2+a^4)geq(1+a+a^2)^2$$ for all $ainmathbb R$.
Now , I solved it , but the problem is that in the answer they wrote this:
using Chebyshev inequality:
$$(1+a+a^2)^2=(1·1+a·1+a^2·1)^2≤(1+a^2+a^4)·(1+1+1)=3(1+a^2+a^4).$$
And so I tried searching the web for this inequality but all it found was the Chebyshev's inequality for probabillity.
can someone please send me link regarding this inequality or just write it here?
Thank you.
algebra-precalculus inequality symmetric-polynomials rearrangement-inequality
edited Jan 1 at 1:24
Michael Rozenberg
97.5k1589188
asked Dec 31 '18 at 23:59
orel Zemachorel Zemach
232
add a comment |
I had a question in my exam and they asked to prove that
prove that:
$$3(1+a^2+a^4)geq(1+a+a^2)^2$$ for all $ainmathbb R$.
Now , I solved it , but the problem is that in the answer they wrote this:
using Chebyshev inequality:
$$(1+a+a^2)^2=(1·1+a·1+a^2·1)^2≤(1+a^2+a^4)·(1+1+1)=3(1+a^2+a^4).$$
And so I tried searching the web for this inequality but all it found was the Chebyshev's inequality for probabillity.
can someone please send me link regarding this inequality or just write it here?
Thank you.
algebra-precalculus inequality symmetric-polynomials rearrangement-inequality
edited Jan 1 at 1:24
Michael Rozenberg
97.5k1589188
asked Dec 31 '18 at 23:59
orel Zemachorel Zemach
232
I had a question in my exam and they asked to prove that
prove that:
$$3(1+a^2+a^4)geq(1+a+a^2)^2$$ for all $ainmathbb R$.
Now , I solved it , but the problem is that in the answer they wrote this:
using Chebyshev inequality:
$$(1+a+a^2)^2=(1·1+a·1+a^2·1)^2≤(1+a^2+a^4)·(1+1+1)=3(1+a^2+a^4).$$
And so I tried searching the web for this inequality but all it found was the Chebyshev's inequality for probabillity.
can someone please send me link regarding this inequality or just write it here?
Thank you.
algebra-precalculus inequality symmetric-polynomials rearrangement-inequality
algebra-precalculus inequality symmetric-polynomials rearrangement-inequality
edited Jan 1 at 1:24
Michael Rozenberg
97.5k1589188
asked Dec 31 '18 at 23:59
orel Zemachorel Zemach
232
edited Jan 1 at 1:24
Michael Rozenberg
97.5k1589188
asked Dec 31 '18 at 23:59
orel Zemachorel Zemach
232
edited Jan 1 at 1:24
Michael Rozenberg
97.5k1589188
edited Jan 1 at 1:24
Michael Rozenberg
97.5k1589188
edited Jan 1 at 1:24
Michael Rozenberg
97.5k1589188
97.5k1589188
asked Dec 31 '18 at 23:59
orel Zemachorel Zemach
232
asked Dec 31 '18 at 23:59
orel Zemachorel Zemach
232
asked Dec 31 '18 at 23:59
orel Zemachorel Zemach
232
232
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
It's just C-S:
$$3(1+a^2+a^4)=(1+1+1)(1+a^2+a^4)geq(1+a+a^2)^2.$$
The Chebyshov inequality it's the following.
Let $a_1geq a_2geq...geq a_n$ and $b_1geq b_2geq...geq b_n$.
Thus, $$nsum_{k=1}^na_kb_{n-k+1}leqsum_{k=1}^na_ksum_{k=1}^nb_kleq nsum_{k=1}^na_kb_k.$$
The proof of this inequality follows immediately from Rearrangement.
There is also the following way:
$$(1+a^2+a^4)=(1+2a^2+a^4-a^2)=((1+a^2)^2-a^2)=(1-a+a^2)(1+a+a^2).$$
Id est, it's enough to prove that:
$$3(1-a+a^2)geq1+a+a^2$$ or
$$(a-1)^2geq0.$$
answered Jan 1 at 0:48
Michael RozenbergMichael Rozenberg
97.5k1589188
add a comment |
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%2f3058135%2fchebyshevs-inequality-question%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It's just C-S:
$$3(1+a^2+a^4)=(1+1+1)(1+a^2+a^4)geq(1+a+a^2)^2.$$
The Chebyshov inequality it's the following.
Let $a_1geq a_2geq...geq a_n$ and $b_1geq b_2geq...geq b_n$.
Thus, $$nsum_{k=1}^na_kb_{n-k+1}leqsum_{k=1}^na_ksum_{k=1}^nb_kleq nsum_{k=1}^na_kb_k.$$
The proof of this inequality follows immediately from Rearrangement.
There is also the following way:
$$(1+a^2+a^4)=(1+2a^2+a^4-a^2)=((1+a^2)^2-a^2)=(1-a+a^2)(1+a+a^2).$$
Id est, it's enough to prove that:
$$3(1-a+a^2)geq1+a+a^2$$ or
$$(a-1)^2geq0.$$
answered Jan 1 at 0:48
Michael RozenbergMichael Rozenberg
97.5k1589188
add a comment |
It's just C-S:
$$3(1+a^2+a^4)=(1+1+1)(1+a^2+a^4)geq(1+a+a^2)^2.$$
The Chebyshov inequality it's the following.
Let $a_1geq a_2geq...geq a_n$ and $b_1geq b_2geq...geq b_n$.
Thus, $$nsum_{k=1}^na_kb_{n-k+1}leqsum_{k=1}^na_ksum_{k=1}^nb_kleq nsum_{k=1}^na_kb_k.$$
The proof of this inequality follows immediately from Rearrangement.
There is also the following way:
$$(1+a^2+a^4)=(1+2a^2+a^4-a^2)=((1+a^2)^2-a^2)=(1-a+a^2)(1+a+a^2).$$
Id est, it's enough to prove that:
$$3(1-a+a^2)geq1+a+a^2$$ or
$$(a-1)^2geq0.$$
answered Jan 1 at 0:48
Michael RozenbergMichael Rozenberg
97.5k1589188
add a comment |
It's just C-S:
$$3(1+a^2+a^4)=(1+1+1)(1+a^2+a^4)geq(1+a+a^2)^2.$$
The Chebyshov inequality it's the following.
Let $a_1geq a_2geq...geq a_n$ and $b_1geq b_2geq...geq b_n$.
Thus, $$nsum_{k=1}^na_kb_{n-k+1}leqsum_{k=1}^na_ksum_{k=1}^nb_kleq nsum_{k=1}^na_kb_k.$$
The proof of this inequality follows immediately from Rearrangement.
There is also the following way:
$$(1+a^2+a^4)=(1+2a^2+a^4-a^2)=((1+a^2)^2-a^2)=(1-a+a^2)(1+a+a^2).$$
Id est, it's enough to prove that:
$$3(1-a+a^2)geq1+a+a^2$$ or
$$(a-1)^2geq0.$$
answered Jan 1 at 0:48
Michael RozenbergMichael Rozenberg
97.5k1589188
It's just C-S:
$$3(1+a^2+a^4)=(1+1+1)(1+a^2+a^4)geq(1+a+a^2)^2.$$
The Chebyshov inequality it's the following.
Let $a_1geq a_2geq...geq a_n$ and $b_1geq b_2geq...geq b_n$.
Thus, $$nsum_{k=1}^na_kb_{n-k+1}leqsum_{k=1}^na_ksum_{k=1}^nb_kleq nsum_{k=1}^na_kb_k.$$
The proof of this inequality follows immediately from Rearrangement.
There is also the following way:
$$(1+a^2+a^4)=(1+2a^2+a^4-a^2)=((1+a^2)^2-a^2)=(1-a+a^2)(1+a+a^2).$$
Id est, it's enough to prove that:
$$3(1-a+a^2)geq1+a+a^2$$ or
$$(a-1)^2geq0.$$
answered Jan 1 at 0:48
Michael RozenbergMichael Rozenberg
97.5k1589188
edited Jan 1 at 1:20
answered Jan 1 at 0:48
Michael RozenbergMichael Rozenberg
97.5k1589188
answered Jan 1 at 0:48
Michael RozenbergMichael Rozenberg
97.5k1589188
answered Jan 1 at 0:48
Michael RozenbergMichael Rozenberg
97.5k1589188
97.5k1589188
add a comment |
add a comment |
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%2f3058135%2fchebyshevs-inequality-question%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
