Dgdrxrt

Should I join office cleaning event for free?

Why has Russell's definition of numbers using equivalence classes been finally abandoned? ( If it has actually been abandoned).

Japan - Plan around max visa duration

What would the Romans have called "sorcery"?

Why Is Death Allowed In the Matrix?

What do you call something that goes against the spirit of the law, but is legal when interpreting the law to the letter?

"You are your self first supporter", a more proper way to say it

Copycat chess is back

Copenhagen passport control - US citizen

Pronouncing Dictionary.com's W.O.D "vade mecum" in English

How to make payment on the internet without leaving a money trail?

Why is this code 6.5x slower with optimizations enabled?

TGV timetables / schedules?

DOS, create pipe for stdin/stdout of command.com(or 4dos.com) in C or Batch?

Can a German sentence have two subjects?

Why don't electromagnetic waves interact with each other?

Work Breakdown with Tikz

Are tax years 2016 & 2017 back taxes deductible for tax year 2018?

Patience, young "Padovan"

I’m planning on buying a laser printer but concerned about the life cycle of toner in the machine

Why linear maps act like matrix multiplication?

How old can references or sources in a thesis be?

When blogging recipes, how can I support both readers who want the narrative/journey and ones who want the printer-friendly recipe?

What is the offset in a seaplane's hull?

How can I prove this inequality of the form $|f_1g_1-f_2g_2|leq|f_1-f_2|$?

Two inequalitiesHow to prove this matrix boundHow to prove this inequality without using Muirhead's inequality?Inequality with complex sineHow prove Reversing the Arithmetic mean – Geometric mean inequality？Prove that $sinh(cosh(x)) geq cosh(sinh(x))$Prove this inequality with best constant?Prove the inequality using AM-GM only.Problem with cosh and sinhHow prove this similar Hilbert inequality

0

$begingroup$

I am trying to prove this inequality:

$$
left|frac1left(left(x+aright)^2+bright)^frac32cdotfracsinhleft(sqrtleft(x+aright)^2+bright)coshleft(sqrtleft(x+aright)^2+bright)+c
-
frac1left(left(x-aright)^2+bright)^frac32cdotfracsinhleft(sqrtleft(x-aright)^2+bright)coshleft(sqrtleft(x-aright)^2+bright)+cright|
leq
left|frac1left(left(x+aright)^2+bright)^frac32-frac1left(left(x-aright)^2+bright)^frac32right|
$$

for all $x>0$ where $a,b,c>0$ are constants.

This could be more briefly expressed as:

$$left|fleft(a,xright)cdot gleft(a,xright)-fleft(-a,xright)cdot gleft(-a,xright)right|leq left|fleft(a,xright)-fleft(-a,xright)right|,$$

where

$$fleft(a,xright)=frac1left(left(x+aright)^2+bright)^frac32, qquad gleft(a,xright)=fracsinhleft(sqrtleft(x+aright)^2+bright)coshleft(sqrtleft(x+aright)^2+bright)+c.$$

After playing around with graphing software I am fairly convinced this inequality holds, but I can't figure out how to prove it. What makes it particularly tricky (to me) is that while $f(a,x)$ is increasing (in $x$), $g(a,x)$ is decreasing, so there is no general consistency as to which term (in the difference) is greater than the other.

Sorry if the inequality is wrong, or insanely difficult to show—I'm just wondering if there's an easy trick that I'm missing.

inequality

share|cite|improve this question

asked Mar 29 at 21:06

WillGWillG

495310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This might help. Replace sinh and cosh by exp. Good lick.
  $endgroup$
  – herb steinberg
  Mar 29 at 21:15
  

  
  
  
  

add a comment | 

0

$begingroup$

I am trying to prove this inequality:

$$
left|frac1left(left(x+aright)^2+bright)^frac32cdotfracsinhleft(sqrtleft(x+aright)^2+bright)coshleft(sqrtleft(x+aright)^2+bright)+c
-
frac1left(left(x-aright)^2+bright)^frac32cdotfracsinhleft(sqrtleft(x-aright)^2+bright)coshleft(sqrtleft(x-aright)^2+bright)+cright|
leq
left|frac1left(left(x+aright)^2+bright)^frac32-frac1left(left(x-aright)^2+bright)^frac32right|
$$

for all $x>0$ where $a,b,c>0$ are constants.

This could be more briefly expressed as:

$$left|fleft(a,xright)cdot gleft(a,xright)-fleft(-a,xright)cdot gleft(-a,xright)right|leq left|fleft(a,xright)-fleft(-a,xright)right|,$$

where

$$fleft(a,xright)=frac1left(left(x+aright)^2+bright)^frac32, qquad gleft(a,xright)=fracsinhleft(sqrtleft(x+aright)^2+bright)coshleft(sqrtleft(x+aright)^2+bright)+c.$$

After playing around with graphing software I am fairly convinced this inequality holds, but I can't figure out how to prove it. What makes it particularly tricky (to me) is that while $f(a,x)$ is increasing (in $x$), $g(a,x)$ is decreasing, so there is no general consistency as to which term (in the difference) is greater than the other.

Sorry if the inequality is wrong, or insanely difficult to show—I'm just wondering if there's an easy trick that I'm missing.

inequality

share|cite|improve this question

asked Mar 29 at 21:06

WillGWillG

495310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This might help. Replace sinh and cosh by exp. Good lick.
  $endgroup$
  – herb steinberg
  Mar 29 at 21:15
  

  
  
  
  

add a comment | 

$begingroup$

I am trying to prove this inequality:

$$
left|frac1left(left(x+aright)^2+bright)^frac32cdotfracsinhleft(sqrtleft(x+aright)^2+bright)coshleft(sqrtleft(x+aright)^2+bright)+c
-
frac1left(left(x-aright)^2+bright)^frac32cdotfracsinhleft(sqrtleft(x-aright)^2+bright)coshleft(sqrtleft(x-aright)^2+bright)+cright|
leq
left|frac1left(left(x+aright)^2+bright)^frac32-frac1left(left(x-aright)^2+bright)^frac32right|
$$

for all $x>0$ where $a,b,c>0$ are constants.

This could be more briefly expressed as:

$$left|fleft(a,xright)cdot gleft(a,xright)-fleft(-a,xright)cdot gleft(-a,xright)right|leq left|fleft(a,xright)-fleft(-a,xright)right|,$$

where

$$fleft(a,xright)=frac1left(left(x+aright)^2+bright)^frac32, qquad gleft(a,xright)=fracsinhleft(sqrtleft(x+aright)^2+bright)coshleft(sqrtleft(x+aright)^2+bright)+c.$$

After playing around with graphing software I am fairly convinced this inequality holds, but I can't figure out how to prove it. What makes it particularly tricky (to me) is that while $f(a,x)$ is increasing (in $x$), $g(a,x)$ is decreasing, so there is no general consistency as to which term (in the difference) is greater than the other.

Sorry if the inequality is wrong, or insanely difficult to show—I'm just wondering if there's an easy trick that I'm missing.

inequality

share|cite|improve this question

asked Mar 29 at 21:06

WillGWillG

495310

$endgroup$

share|cite|improve this question

asked Mar 29 at 21:06

WillGWillG

495310

share|cite|improve this question

asked Mar 29 at 21:06

WillGWillG

495310

asked Mar 29 at 21:06

WillGWillG

495310

asked Mar 29 at 21:06

WillGWillG

495310

1 Answer
1

active

oldest

votes

1

$begingroup$

Something like this
(which is a standard grouping technique)
might be a start.

Simplifying the notation somewhat,
you want to show that

$|f(a)g(a)-f(-a) g(-a)|
leq |f(a)-f(-a)|,
$

We have

$beginarray\
|f(a)g(a)-f(-a) g(-a)|
&=|f(a)g(a)-f(a)g(-a)+f(a)g(-a)-f(-a) g(-a)|\
&le|f(a)g(a)-f(a)g(-a)|+|f(a)g(-a)-f(-a) g(-a)|\
&le |f(a)||g(a)-g(-a)|+|g(-a)||f(a)-f(-a)|\
endarray
$

In your case,
$|g| < 1$.
This might also hold
for $f$
depending on the values of
$a$ and $b$.

More analysis of
$f$ and $g$
might get useful results,
but I'll stop here.

share|cite|improve this answer

answered Mar 29 at 23:12

marty cohenmarty cohen

75k549130

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Excellent. I'm not sure I can use this to prove my inequality, but I don't need to. My ultimate purpose is to use the dominated convergence theorem to move a limit inside an integral, and this will split the integral into two pieces that can be handled separately. Thanks!
  $endgroup$
  – WillG
  Mar 30 at 0:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Glad to have helped. This sort of thing is the reason that I submit incomplete answers when I haven't completely solved the problem.
  $endgroup$
  – marty cohen
  Mar 30 at 2:22
  

  
  
  
  

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3167617%2fhow-can-i-prove-this-inequality-of-the-form-f-1g-1-f-2g-2-leqf-1-f-2%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

Something like this
(which is a standard grouping technique)
might be a start.

Simplifying the notation somewhat,
you want to show that

$|f(a)g(a)-f(-a) g(-a)|
leq |f(a)-f(-a)|,
$

We have

$beginarray\
|f(a)g(a)-f(-a) g(-a)|
&=|f(a)g(a)-f(a)g(-a)+f(a)g(-a)-f(-a) g(-a)|\
&le|f(a)g(a)-f(a)g(-a)|+|f(a)g(-a)-f(-a) g(-a)|\
&le |f(a)||g(a)-g(-a)|+|g(-a)||f(a)-f(-a)|\
endarray
$

In your case,
$|g| < 1$.
This might also hold
for $f$
depending on the values of
$a$ and $b$.

More analysis of
$f$ and $g$
might get useful results,
but I'll stop here.

share|cite|improve this answer

answered Mar 29 at 23:12

marty cohenmarty cohen

75k549130

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Excellent. I'm not sure I can use this to prove my inequality, but I don't need to. My ultimate purpose is to use the dominated convergence theorem to move a limit inside an integral, and this will split the integral into two pieces that can be handled separately. Thanks!
  $endgroup$
  – WillG
  Mar 30 at 0:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Glad to have helped. This sort of thing is the reason that I submit incomplete answers when I haven't completely solved the problem.
  $endgroup$
  – marty cohen
  Mar 30 at 2:22
  

  
  
  
  

add a comment | 

1

$begingroup$

Something like this
(which is a standard grouping technique)
might be a start.

Simplifying the notation somewhat,
you want to show that

$|f(a)g(a)-f(-a) g(-a)|
leq |f(a)-f(-a)|,
$

We have

$beginarray\
|f(a)g(a)-f(-a) g(-a)|
&=|f(a)g(a)-f(a)g(-a)+f(a)g(-a)-f(-a) g(-a)|\
&le|f(a)g(a)-f(a)g(-a)|+|f(a)g(-a)-f(-a) g(-a)|\
&le |f(a)||g(a)-g(-a)|+|g(-a)||f(a)-f(-a)|\
endarray
$

In your case,
$|g| < 1$.
This might also hold
for $f$
depending on the values of
$a$ and $b$.

More analysis of
$f$ and $g$
might get useful results,
but I'll stop here.

share|cite|improve this answer

answered Mar 29 at 23:12

marty cohenmarty cohen

75k549130

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Excellent. I'm not sure I can use this to prove my inequality, but I don't need to. My ultimate purpose is to use the dominated convergence theorem to move a limit inside an integral, and this will split the integral into two pieces that can be handled separately. Thanks!
  $endgroup$
  – WillG
  Mar 30 at 0:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Glad to have helped. This sort of thing is the reason that I submit incomplete answers when I haven't completely solved the problem.
  $endgroup$
  – marty cohen
  Mar 30 at 2:22
  

  
  
  
  

add a comment | 

$begingroup$

Something like this
(which is a standard grouping technique)
might be a start.

Simplifying the notation somewhat,
you want to show that

$|f(a)g(a)-f(-a) g(-a)|
leq |f(a)-f(-a)|,
$

We have

$beginarray\
|f(a)g(a)-f(-a) g(-a)|
&=|f(a)g(a)-f(a)g(-a)+f(a)g(-a)-f(-a) g(-a)|\
&le|f(a)g(a)-f(a)g(-a)|+|f(a)g(-a)-f(-a) g(-a)|\
&le |f(a)||g(a)-g(-a)|+|g(-a)||f(a)-f(-a)|\
endarray
$

In your case,
$|g| < 1$.
This might also hold
for $f$
depending on the values of
$a$ and $b$.

More analysis of
$f$ and $g$
might get useful results,
but I'll stop here.

share|cite|improve this answer

answered Mar 29 at 23:12

marty cohenmarty cohen

75k549130

$endgroup$

share|cite|improve this answer

answered Mar 29 at 23:12

marty cohenmarty cohen

75k549130

answered Mar 29 at 23:12

marty cohenmarty cohen

75k549130

answered Mar 29 at 23:12

marty cohenmarty cohen

75k549130