Arithmetic mean geometric mean inequality unclearproving inequality?Practicing the arithmetic-geometric means...
Do the temporary hit points from the Battlerager barbarian's Reckless Abandon stack if I make multiple attacks on my turn?
Why are there no referendums in the US?
What can we do to stop prior company from asking us questions?
How do I go from 300 unfinished/half written blog posts, to published posts?
Is exact Kanji stroke length important?
Method to test if a number is a perfect power?
Escape a backup date in a file name
How can I get through very long and very dry, but also very useful technical documents when learning a new tool?
Do sorcerers' Subtle Spells require a skill check to be unseen?
Is a stroke of luck acceptable after a series of unfavorable events?
Would a high gravity rocky planet be guaranteed to have an atmosphere?
Is the destination of a commercial flight important for the pilot?
Sequence of Tenses: Translating the subjunctive
Why Were Madagascar and New Zealand Discovered So Late?
Failed to fetch jessie backports repository
How easy is it to start Magic from scratch?
Trouble understanding the speech of overseas colleagues
Roman Numeral Treatment of Suspensions
Crossing the line between justified force and brutality
Pre-amplifier input protection
Is there a good way to store credentials outside of a password manager?
Term for the "extreme-extension" version of a straw man fallacy?
A Rare Riley Riddle
How to check is there any negative term in a large list?
Arithmetic mean geometric mean inequality unclear
proving inequality?Practicing the arithmetic-geometric means inequalityArithmetic Mean and Geometric Mean Question, Guidance NeededHow prove Reversing the Arithmetic mean – Geometric mean inequality？Mean Value Theorem and Inequality.Using arithmetic mean>geometric meanNesbitt's Inequality $frac{a}{b+c}+frac{b}{c+a}+frac{c}{a+b}geqfrac{3}{2}$Problem in Arithmetic Mean - Geometric Mean inequalityProving Cauchy-Schwarz with Arithmetic Geometric meanInequality involving a kind of Harmonic mean
$begingroup$
I know that the AM-GM inequality takes the form $$ frac{x + y}{2} geq sqrt{xy},$$ but I read in a book another form which is $$ frac{x^2 + y^2}{2} geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me please?
calculus inequality
edited 5 hours ago
Bernard
123k741117
asked 5 hours ago
hopefullyhopefully
274114
$endgroup$
add a comment |
$begingroup$
I know that the AM-GM inequality takes the form $$ frac{x + y}{2} geq sqrt{xy},$$ but I read in a book another form which is $$ frac{x^2 + y^2}{2} geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me please?
calculus inequality
edited 5 hours ago
Bernard
123k741117
asked 5 hours ago
hopefullyhopefully
274114
$endgroup$
add a comment |
$begingroup$
I know that the AM-GM inequality takes the form $$ frac{x + y}{2} geq sqrt{xy},$$ but I read in a book another form which is $$ frac{x^2 + y^2}{2} geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me please?
calculus inequality
edited 5 hours ago
Bernard
123k741117
asked 5 hours ago
hopefullyhopefully
274114
$endgroup$
I know that the AM-GM inequality takes the form $$ frac{x + y}{2} geq sqrt{xy},$$ but I read in a book another form which is $$ frac{x^2 + y^2}{2} geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me please?
calculus inequality
calculus inequality
edited 5 hours ago
Bernard
123k741117
asked 5 hours ago
hopefullyhopefully
274114
edited 5 hours ago
Bernard
123k741117
asked 5 hours ago
hopefullyhopefully
274114
edited 5 hours ago
Bernard
123k741117
edited 5 hours ago
Bernard
123k741117
edited 5 hours ago
Bernard
123k741117
123k741117
asked 5 hours ago
hopefullyhopefully
274114
asked 5 hours ago
hopefullyhopefully
274114
asked 5 hours ago
hopefullyhopefully
274114
274114
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
$$frac{X^2+Y^2}{2} ge sqrt{X^2Y^2} = sqrt{(XY)^2}=|XY|,$$
which is the second inequality (modulo capitalization).
answered 5 hours ago
jgonjgon
16k32143
$endgroup$
add a comment |
$begingroup$
The AM-GM inequality for $n$ non-negative values is
$frac1{n}(sum_{k=1}^n x_k)
ge (prod_{k=1}^n x_k)^{1/n}
$.
This can be rewritten in two ways.
First,
by simple algebra,
$(sum_{k=1}^n x_i)^n
ge n^n(prod_{k=1}^n x_k)
$.
Second,
letting $x_k = y_k^n$,
this becomes
$frac1{n}(sum_{k=1}^n y_k^n)
ge prod_{k=1}^n y_k
$.
It is useful to recognize
these disguises.
answered 5 hours ago
marty cohenmarty cohen
74.9k549130
$endgroup$
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%2f3165273%2farithmetic-mean-geometric-mean-inequality-unclear%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
$$frac{X^2+Y^2}{2} ge sqrt{X^2Y^2} = sqrt{(XY)^2}=|XY|,$$
which is the second inequality (modulo capitalization).
answered 5 hours ago
jgonjgon
16k32143
$endgroup$
add a comment |
$begingroup$
If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
$$frac{X^2+Y^2}{2} ge sqrt{X^2Y^2} = sqrt{(XY)^2}=|XY|,$$
which is the second inequality (modulo capitalization).
answered 5 hours ago
jgonjgon
16k32143
$endgroup$
add a comment |
$begingroup$
If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
$$frac{X^2+Y^2}{2} ge sqrt{X^2Y^2} = sqrt{(XY)^2}=|XY|,$$
which is the second inequality (modulo capitalization).
answered 5 hours ago
jgonjgon
16k32143
$endgroup$
If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
$$frac{X^2+Y^2}{2} ge sqrt{X^2Y^2} = sqrt{(XY)^2}=|XY|,$$
which is the second inequality (modulo capitalization).
answered 5 hours ago
jgonjgon
16k32143
answered 5 hours ago
jgonjgon
16k32143
answered 5 hours ago
jgonjgon
16k32143
answered 5 hours ago
jgonjgon
16k32143
16k32143
add a comment |
add a comment |
$begingroup$
The AM-GM inequality for $n$ non-negative values is
$frac1{n}(sum_{k=1}^n x_k)
ge (prod_{k=1}^n x_k)^{1/n}
$.
This can be rewritten in two ways.
First,
by simple algebra,
$(sum_{k=1}^n x_i)^n
ge n^n(prod_{k=1}^n x_k)
$.
Second,
letting $x_k = y_k^n$,
this becomes
$frac1{n}(sum_{k=1}^n y_k^n)
ge prod_{k=1}^n y_k
$.
It is useful to recognize
these disguises.
answered 5 hours ago
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
The AM-GM inequality for $n$ non-negative values is
$frac1{n}(sum_{k=1}^n x_k)
ge (prod_{k=1}^n x_k)^{1/n}
$.
This can be rewritten in two ways.
First,
by simple algebra,
$(sum_{k=1}^n x_i)^n
ge n^n(prod_{k=1}^n x_k)
$.
Second,
letting $x_k = y_k^n$,
this becomes
$frac1{n}(sum_{k=1}^n y_k^n)
ge prod_{k=1}^n y_k
$.
It is useful to recognize
these disguises.
answered 5 hours ago
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
The AM-GM inequality for $n$ non-negative values is
$frac1{n}(sum_{k=1}^n x_k)
ge (prod_{k=1}^n x_k)^{1/n}
$.
This can be rewritten in two ways.
First,
by simple algebra,
$(sum_{k=1}^n x_i)^n
ge n^n(prod_{k=1}^n x_k)
$.
Second,
letting $x_k = y_k^n$,
this becomes
$frac1{n}(sum_{k=1}^n y_k^n)
ge prod_{k=1}^n y_k
$.
It is useful to recognize
these disguises.
answered 5 hours ago
marty cohenmarty cohen
74.9k549130
$endgroup$
The AM-GM inequality for $n$ non-negative values is
$frac1{n}(sum_{k=1}^n x_k)
ge (prod_{k=1}^n x_k)^{1/n}
$.
This can be rewritten in two ways.
First,
by simple algebra,
$(sum_{k=1}^n x_i)^n
ge n^n(prod_{k=1}^n x_k)
$.
Second,
letting $x_k = y_k^n$,
this becomes
$frac1{n}(sum_{k=1}^n y_k^n)
ge prod_{k=1}^n y_k
$.
It is useful to recognize
these disguises.
answered 5 hours ago
marty cohenmarty cohen
74.9k549130
answered 5 hours ago
marty cohenmarty cohen
74.9k549130
answered 5 hours ago
marty cohenmarty cohen
74.9k549130
answered 5 hours ago
marty cohenmarty cohen
74.9k549130
74.9k549130
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%2f3165273%2farithmetic-mean-geometric-mean-inequality-unclear%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
