Mixing
Another Stubborn Inequality
$begingroup$
ReBonjour.
Let $x$ and $y$ be two real numbers. Show that $ frac{ |x+y|}{1+|x+y|}leqfrac{|x|}{1+|x|}+frac {|y|}{1+|y|}$
Thanks a lot.
real-analysis calculus geometry analysis
asked Jan 13 at 16:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
73229
$endgroup$
add a comment |
$begingroup$
ReBonjour.
Let $x$ and $y$ be two real numbers. Show that $ frac{ |x+y|}{1+|x+y|}leqfrac{|x|}{1+|x|}+frac {|y|}{1+|y|}$
Thanks a lot.
real-analysis calculus geometry analysis
asked Jan 13 at 16:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
73229
$endgroup$
$begingroup$
If both $x$ and $y$ are zero, then this is false. I think you mean $frac{ |x+y|}{1+|x+y|}leqfrac{|x|}{1+|x|}+frac {|y|}{1+|y|}$
$endgroup$
– R. Burton
Jan 13 at 20:51
$begingroup$
Corrected. Thanks
$endgroup$
– HAMIDINE SOUMARE
Jan 13 at 20:53
add a comment |
$begingroup$
ReBonjour.
Let $x$ and $y$ be two real numbers. Show that $ frac{ |x+y|}{1+|x+y|}leqfrac{|x|}{1+|x|}+frac {|y|}{1+|y|}$
Thanks a lot.
real-analysis calculus geometry analysis
asked Jan 13 at 16:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
73229
$endgroup$
ReBonjour.
Let $x$ and $y$ be two real numbers. Show that $ frac{ |x+y|}{1+|x+y|}leqfrac{|x|}{1+|x|}+frac {|y|}{1+|y|}$
Thanks a lot.
real-analysis calculus geometry analysis
real-analysis calculus geometry analysis
asked Jan 13 at 16:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
73229
asked Jan 13 at 16:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
73229
edited Jan 13 at 20:52
HAMIDINE SOUMARE
asked Jan 13 at 16:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
73229
asked Jan 13 at 16:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
73229
asked Jan 13 at 16:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
73229
73229
$begingroup$
If both $x$ and $y$ are zero, then this is false. I think you mean $frac{ |x+y|}{1+|x+y|}leqfrac{|x|}{1+|x|}+frac {|y|}{1+|y|}$
$endgroup$
– R. Burton
Jan 13 at 20:51
$begingroup$
Corrected. Thanks
$endgroup$
– HAMIDINE SOUMARE
Jan 13 at 20:53
add a comment |
$begingroup$
If both $x$ and $y$ are zero, then this is false. I think you mean $frac{ |x+y|}{1+|x+y|}leqfrac{|x|}{1+|x|}+frac {|y|}{1+|y|}$
$endgroup$
– R. Burton
Jan 13 at 20:51
$begingroup$
Corrected. Thanks
$endgroup$
– HAMIDINE SOUMARE
Jan 13 at 20:53
$begingroup$
If both $x$ and $y$ are zero, then this is false. I think you mean $frac{ |x+y|}{1+|x+y|}leqfrac{|x|}{1+|x|}+frac {|y|}{1+|y|}$
$endgroup$
– R. Burton
Jan 13 at 20:51
$begingroup$
If both $x$ and $y$ are zero, then this is false. I think you mean $frac{ |x+y|}{1+|x+y|}leqfrac{|x|}{1+|x|}+frac {|y|}{1+|y|}$
$endgroup$
– R. Burton
Jan 13 at 20:51
$begingroup$
Corrected. Thanks
$endgroup$
– HAMIDINE SOUMARE
Jan 13 at 20:53
$begingroup$
Corrected. Thanks
$endgroup$
– HAMIDINE SOUMARE
Jan 13 at 20:53
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that the function
$$
f(x)=frac{x}{1+x}
$$
is increasing (by taking derivatives or simply by using the fact that $frac{a}{b}<frac{c}{d}iff ad<bc$) whence
$$
|x+y|leq|x|+|y|
$$
implies that
$$
begin{align}
frac{|x+y|}{1+|x+y|}leqfrac{|x|+|y|}{1+|x|+|y|}&=frac{|x|}{1+|x|+|y|}+frac{|y|}{1+|x|+|y|}\&leq
frac{|x|}{1+|x|}+
frac{|y|}{1+|y|}
end{align}
$$
as desired.
answered Jan 13 at 16:36
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
add a comment |
$begingroup$
Observe that $$1-frac{1}{1+|x+y|}leq 1-frac{1}{1+|x|}$$ $$1-frac{1}{1+|x+y|}leq 1-frac{1}{1+|y|}$$
So $$2-frac{2}{1+|x+y|}leq 1-frac{1}{1+|x|}+1-frac{1}{1+|y|}$$
Thus $$frac{2|x+y|}{1+|x+y|}leq frac{|x|}{1+|x|}+frac{|y|}{1+|y|}$$
$$frac{|x+y|}{1+|x+y|}< frac{|x|}{1+|x|}+frac{|y|}{1+|y|}$$
answered Jan 13 at 16:25
Angelo MarkAngelo Mark
4,03721641
$endgroup$
$begingroup$
If $x=y=0$, strict inequality does not occur.
$endgroup$
– Aweygan
Jan 13 at 16:26
$begingroup$
Angelo $x$ is not supposed to be positive. Your argument does not hold.
$endgroup$
– HAMIDINE SOUMARE
Jan 13 at 16:44
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%2f3072168%2fanother-stubborn-inequality%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
$begingroup$
Note that the function
$$
f(x)=frac{x}{1+x}
$$
is increasing (by taking derivatives or simply by using the fact that $frac{a}{b}<frac{c}{d}iff ad<bc$) whence
$$
|x+y|leq|x|+|y|
$$
implies that
$$
begin{align}
frac{|x+y|}{1+|x+y|}leqfrac{|x|+|y|}{1+|x|+|y|}&=frac{|x|}{1+|x|+|y|}+frac{|y|}{1+|x|+|y|}\&leq
frac{|x|}{1+|x|}+
frac{|y|}{1+|y|}
end{align}
$$
as desired.
answered Jan 13 at 16:36
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
add a comment |
$begingroup$
Note that the function
$$
f(x)=frac{x}{1+x}
$$
is increasing (by taking derivatives or simply by using the fact that $frac{a}{b}<frac{c}{d}iff ad<bc$) whence
$$
|x+y|leq|x|+|y|
$$
implies that
$$
begin{align}
frac{|x+y|}{1+|x+y|}leqfrac{|x|+|y|}{1+|x|+|y|}&=frac{|x|}{1+|x|+|y|}+frac{|y|}{1+|x|+|y|}\&leq
frac{|x|}{1+|x|}+
frac{|y|}{1+|y|}
end{align}
$$
as desired.
answered Jan 13 at 16:36
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
add a comment |
$begingroup$
Note that the function
$$
f(x)=frac{x}{1+x}
$$
is increasing (by taking derivatives or simply by using the fact that $frac{a}{b}<frac{c}{d}iff ad<bc$) whence
$$
|x+y|leq|x|+|y|
$$
implies that
$$
begin{align}
frac{|x+y|}{1+|x+y|}leqfrac{|x|+|y|}{1+|x|+|y|}&=frac{|x|}{1+|x|+|y|}+frac{|y|}{1+|x|+|y|}\&leq
frac{|x|}{1+|x|}+
frac{|y|}{1+|y|}
end{align}
$$
as desired.
answered Jan 13 at 16:36
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
Note that the function
$$
f(x)=frac{x}{1+x}
$$
is increasing (by taking derivatives or simply by using the fact that $frac{a}{b}<frac{c}{d}iff ad<bc$) whence
$$
|x+y|leq|x|+|y|
$$
implies that
$$
begin{align}
frac{|x+y|}{1+|x+y|}leqfrac{|x|+|y|}{1+|x|+|y|}&=frac{|x|}{1+|x|+|y|}+frac{|y|}{1+|x|+|y|}\&leq
frac{|x|}{1+|x|}+
frac{|y|}{1+|y|}
end{align}
$$
as desired.
answered Jan 13 at 16:36
Foobaz JohnFoobaz John
22.1k41352
answered Jan 13 at 16:36
Foobaz JohnFoobaz John
22.1k41352
answered Jan 13 at 16:36
Foobaz JohnFoobaz John
22.1k41352
answered Jan 13 at 16:36
Foobaz JohnFoobaz John
22.1k41352
22.1k41352
add a comment |
add a comment |
$begingroup$
Observe that $$1-frac{1}{1+|x+y|}leq 1-frac{1}{1+|x|}$$ $$1-frac{1}{1+|x+y|}leq 1-frac{1}{1+|y|}$$
So $$2-frac{2}{1+|x+y|}leq 1-frac{1}{1+|x|}+1-frac{1}{1+|y|}$$
Thus $$frac{2|x+y|}{1+|x+y|}leq frac{|x|}{1+|x|}+frac{|y|}{1+|y|}$$
$$frac{|x+y|}{1+|x+y|}< frac{|x|}{1+|x|}+frac{|y|}{1+|y|}$$
answered Jan 13 at 16:25
Angelo MarkAngelo Mark
4,03721641
$endgroup$
$begingroup$
If $x=y=0$, strict inequality does not occur.
$endgroup$
– Aweygan
Jan 13 at 16:26
$begingroup$
Angelo $x$ is not supposed to be positive. Your argument does not hold.
$endgroup$
– HAMIDINE SOUMARE
Jan 13 at 16:44
add a comment |
$begingroup$
Observe that $$1-frac{1}{1+|x+y|}leq 1-frac{1}{1+|x|}$$ $$1-frac{1}{1+|x+y|}leq 1-frac{1}{1+|y|}$$
So $$2-frac{2}{1+|x+y|}leq 1-frac{1}{1+|x|}+1-frac{1}{1+|y|}$$
Thus $$frac{2|x+y|}{1+|x+y|}leq frac{|x|}{1+|x|}+frac{|y|}{1+|y|}$$
$$frac{|x+y|}{1+|x+y|}< frac{|x|}{1+|x|}+frac{|y|}{1+|y|}$$
answered Jan 13 at 16:25
Angelo MarkAngelo Mark
4,03721641
$endgroup$
$begingroup$
If $x=y=0$, strict inequality does not occur.
$endgroup$
– Aweygan
Jan 13 at 16:26
$begingroup$
Angelo $x$ is not supposed to be positive. Your argument does not hold.
$endgroup$
– HAMIDINE SOUMARE
Jan 13 at 16:44
add a comment |
$begingroup$
Observe that $$1-frac{1}{1+|x+y|}leq 1-frac{1}{1+|x|}$$ $$1-frac{1}{1+|x+y|}leq 1-frac{1}{1+|y|}$$
So $$2-frac{2}{1+|x+y|}leq 1-frac{1}{1+|x|}+1-frac{1}{1+|y|}$$
Thus $$frac{2|x+y|}{1+|x+y|}leq frac{|x|}{1+|x|}+frac{|y|}{1+|y|}$$
$$frac{|x+y|}{1+|x+y|}< frac{|x|}{1+|x|}+frac{|y|}{1+|y|}$$
answered Jan 13 at 16:25
Angelo MarkAngelo Mark
4,03721641
$endgroup$
Observe that $$1-frac{1}{1+|x+y|}leq 1-frac{1}{1+|x|}$$ $$1-frac{1}{1+|x+y|}leq 1-frac{1}{1+|y|}$$
So $$2-frac{2}{1+|x+y|}leq 1-frac{1}{1+|x|}+1-frac{1}{1+|y|}$$
Thus $$frac{2|x+y|}{1+|x+y|}leq frac{|x|}{1+|x|}+frac{|y|}{1+|y|}$$
$$frac{|x+y|}{1+|x+y|}< frac{|x|}{1+|x|}+frac{|y|}{1+|y|}$$
answered Jan 13 at 16:25
Angelo MarkAngelo Mark
4,03721641
edited Jan 13 at 16:30
answered Jan 13 at 16:25
Angelo MarkAngelo Mark
4,03721641
answered Jan 13 at 16:25
Angelo MarkAngelo Mark
4,03721641
answered Jan 13 at 16:25
Angelo MarkAngelo Mark
4,03721641
4,03721641
$begingroup$
If $x=y=0$, strict inequality does not occur.
$endgroup$
– Aweygan
Jan 13 at 16:26
$begingroup$
Angelo $x$ is not supposed to be positive. Your argument does not hold.
$endgroup$
– HAMIDINE SOUMARE
Jan 13 at 16:44
add a comment |
$begingroup$
If $x=y=0$, strict inequality does not occur.
$endgroup$
– Aweygan
Jan 13 at 16:26
$begingroup$
Angelo $x$ is not supposed to be positive. Your argument does not hold.
$endgroup$
– HAMIDINE SOUMARE
Jan 13 at 16:44
$begingroup$
If $x=y=0$, strict inequality does not occur.
$endgroup$
– Aweygan
Jan 13 at 16:26
$begingroup$
If $x=y=0$, strict inequality does not occur.
$endgroup$
– Aweygan
Jan 13 at 16:26
$begingroup$
Angelo $x$ is not supposed to be positive. Your argument does not hold.
$endgroup$
– HAMIDINE SOUMARE
Jan 13 at 16:44
$begingroup$
Angelo $x$ is not supposed to be positive. Your argument does not hold.
$endgroup$
– HAMIDINE SOUMARE
Jan 13 at 16:44
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%2f3072168%2fanother-stubborn-inequality%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
$begingroup$
If both $x$ and $y$ are zero, then this is false. I think you mean $frac{ |x+y|}{1+|x+y|}leqfrac{|x|}{1+|x|}+frac {|y|}{1+|y|}$
$endgroup$
– R. Burton
Jan 13 at 20:51
$begingroup$
Corrected. Thanks
$endgroup$
– HAMIDINE SOUMARE
Jan 13 at 20:53