Arithmetic mean geometric mean inequality unclear
$begingroup$
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');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
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');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
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');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
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');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
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');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
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
