Show $sqrt{x (1-y)} - sqrt{y (1-x)} + ysqrt{x} - xsqrt{y} - frac{x - y}{3} > 0$
up vote
1
down vote
favorite
Consider the inequality
$$
sqrt{x (1-y)} - sqrt{y (1-x)} + ysqrt{x} - xsqrt{y} - frac{x - y}{3} > 0
$$
where $0<y<x<1$ and $x+yleq 1$.
One can, for instance, consider optimizing the expression using the Karush-Kuhn-Tucker conditions. However, the computations involved are tedious and unwieldy.
Can the inequality be shown in a simpler way?
multivariable-calculus inequality
edited Dec 6 at 17:52
Glorfindel
3,41981730
asked Dec 5 at 18:46
broncoAbierto
20119
add a comment |
up vote
1
down vote
favorite
Consider the inequality
$$
sqrt{x (1-y)} - sqrt{y (1-x)} + ysqrt{x} - xsqrt{y} - frac{x - y}{3} > 0
$$
where $0<y<x<1$ and $x+yleq 1$.
One can, for instance, consider optimizing the expression using the Karush-Kuhn-Tucker conditions. However, the computations involved are tedious and unwieldy.
Can the inequality be shown in a simpler way?
multivariable-calculus inequality
edited Dec 6 at 17:52
Glorfindel
3,41981730
asked Dec 5 at 18:46
broncoAbierto
20119
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the inequality
$$
sqrt{x (1-y)} - sqrt{y (1-x)} + ysqrt{x} - xsqrt{y} - frac{x - y}{3} > 0
$$
where $0<y<x<1$ and $x+yleq 1$.
One can, for instance, consider optimizing the expression using the Karush-Kuhn-Tucker conditions. However, the computations involved are tedious and unwieldy.
Can the inequality be shown in a simpler way?
multivariable-calculus inequality
edited Dec 6 at 17:52
Glorfindel
3,41981730
asked Dec 5 at 18:46
broncoAbierto
20119
Consider the inequality
$$
sqrt{x (1-y)} - sqrt{y (1-x)} + ysqrt{x} - xsqrt{y} - frac{x - y}{3} > 0
$$
where $0<y<x<1$ and $x+yleq 1$.
One can, for instance, consider optimizing the expression using the Karush-Kuhn-Tucker conditions. However, the computations involved are tedious and unwieldy.
Can the inequality be shown in a simpler way?
multivariable-calculus inequality
multivariable-calculus inequality
edited Dec 6 at 17:52
Glorfindel
3,41981730
asked Dec 5 at 18:46
broncoAbierto
20119
edited Dec 6 at 17:52
Glorfindel
3,41981730
asked Dec 5 at 18:46
broncoAbierto
20119
edited Dec 6 at 17:52
Glorfindel
3,41981730
edited Dec 6 at 17:52
Glorfindel
3,41981730
edited Dec 6 at 17:52
Glorfindel
3,41981730
3,41981730
asked Dec 5 at 18:46
broncoAbierto
20119
asked Dec 5 at 18:46
broncoAbierto
20119
asked Dec 5 at 18:46
broncoAbierto
20119
20119
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
If you let $x = sin^2 a, y = sin^2 b$ with $0<b<a<pi/2$, then your inequality is equivalent to:
$$sin acos b - sin bcos a +sin^2bsin a - sin^2asin b >dfrac{sin^2 a-sin^2 b}{3}$$ or $$sin(a-b)>(sin a-sin b)left(frac 13+sin asin bright)$$
or $$dfrac{cosleft(frac {a-b}2right)}{cosleft(frac{a+b}{2}right)}>frac{1}{3}+frac 12(cos(a-b)-cos(a+b)).$$
Here, the remaining requirement is the same as $sin aleqcos b = sin(pi/2-b)$, which is the same as: $$a+bleqdfrac{pi}{2}.$$
Now let $1>cosleft(frac{a-b}{2}right) = s>cosleft(frac{a+b}{2}right)=tgeqdfrac{1}{sqrt{2}}.$ Then, the equivalent inequality is:
$$frac st>frac 13+frac 12(2s^2-1 - 2t^2+1) = frac 13+s^2-t^2$$
But this one is trivial as:
$$frac 13+s^2-t^2<frac 13+1-frac 12 = frac 56 <1<frac st.$$
answered Dec 6 at 0:37
dezdichado
6,0991929
add a comment |
up vote
1
down vote
For a traditional kind of proof I use beside $,0<y<x<1,$ and $,x+yleq 1,$ also
first derivation ($:=0$) and second derivation ($neq 0$) of a term to get minimum and maximum.
Equivalent to the given inequation is $enspacedisplaystyle sqrt{x(1-y)}-sqrt{y(1-x)} > xsqrt{y}- ysqrt{x} + frac{x-y}{3}$
and we devide by $,x-y,$ to get $enspacedisplaystyle frac{1}{sqrt{x(1-y)}+sqrt{y(1-x)}} > frac{1}{frac{1}{sqrt{y}}+frac{1}{sqrt{x}}} + frac{1}{3} ,$ .
Using first and second derivation we get
$displaystyle sqrt{x(1-y)}+sqrt{y(1-x)} leq max(sqrt{x(1-y)}+sqrt{y(1-x)}) = 1 enspace$ and
$displaystyle frac{1}{sqrt{y}}+frac{1}{sqrt{x}} geq frac{1}{sqrt{1-x}}+frac{1}{sqrt{x}} geq min(frac{1}{sqrt{1-x}}+frac{1}{sqrt{x}}) = 2sqrt{2}enspace$ so that follows
$displaystyle frac{1}{sqrt{x(1-y)}+sqrt{y(1-x)}} geq 1 > frac{1}{2sqrt{2}} + frac{1}{3} geq frac{1}{frac{1}{sqrt{y}}+frac{1}{sqrt{x}}} + frac{1}{3} enspace$ as wished.
answered Dec 13 at 11:30
user90369
8,193925
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%2f3027476%2fshow-sqrtx-1-y-sqrty-1-x-y-sqrtx-x-sqrty-fracx-y3%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
up vote
3
down vote
accepted
If you let $x = sin^2 a, y = sin^2 b$ with $0<b<a<pi/2$, then your inequality is equivalent to:
$$sin acos b - sin bcos a +sin^2bsin a - sin^2asin b >dfrac{sin^2 a-sin^2 b}{3}$$ or $$sin(a-b)>(sin a-sin b)left(frac 13+sin asin bright)$$
or $$dfrac{cosleft(frac {a-b}2right)}{cosleft(frac{a+b}{2}right)}>frac{1}{3}+frac 12(cos(a-b)-cos(a+b)).$$
Here, the remaining requirement is the same as $sin aleqcos b = sin(pi/2-b)$, which is the same as: $$a+bleqdfrac{pi}{2}.$$
Now let $1>cosleft(frac{a-b}{2}right) = s>cosleft(frac{a+b}{2}right)=tgeqdfrac{1}{sqrt{2}}.$ Then, the equivalent inequality is:
$$frac st>frac 13+frac 12(2s^2-1 - 2t^2+1) = frac 13+s^2-t^2$$
But this one is trivial as:
$$frac 13+s^2-t^2<frac 13+1-frac 12 = frac 56 <1<frac st.$$
answered Dec 6 at 0:37
dezdichado
6,0991929
add a comment |
up vote
3
down vote
accepted
If you let $x = sin^2 a, y = sin^2 b$ with $0<b<a<pi/2$, then your inequality is equivalent to:
$$sin acos b - sin bcos a +sin^2bsin a - sin^2asin b >dfrac{sin^2 a-sin^2 b}{3}$$ or $$sin(a-b)>(sin a-sin b)left(frac 13+sin asin bright)$$
or $$dfrac{cosleft(frac {a-b}2right)}{cosleft(frac{a+b}{2}right)}>frac{1}{3}+frac 12(cos(a-b)-cos(a+b)).$$
Here, the remaining requirement is the same as $sin aleqcos b = sin(pi/2-b)$, which is the same as: $$a+bleqdfrac{pi}{2}.$$
Now let $1>cosleft(frac{a-b}{2}right) = s>cosleft(frac{a+b}{2}right)=tgeqdfrac{1}{sqrt{2}}.$ Then, the equivalent inequality is:
$$frac st>frac 13+frac 12(2s^2-1 - 2t^2+1) = frac 13+s^2-t^2$$
But this one is trivial as:
$$frac 13+s^2-t^2<frac 13+1-frac 12 = frac 56 <1<frac st.$$
answered Dec 6 at 0:37
dezdichado
6,0991929
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
If you let $x = sin^2 a, y = sin^2 b$ with $0<b<a<pi/2$, then your inequality is equivalent to:
$$sin acos b - sin bcos a +sin^2bsin a - sin^2asin b >dfrac{sin^2 a-sin^2 b}{3}$$ or $$sin(a-b)>(sin a-sin b)left(frac 13+sin asin bright)$$
or $$dfrac{cosleft(frac {a-b}2right)}{cosleft(frac{a+b}{2}right)}>frac{1}{3}+frac 12(cos(a-b)-cos(a+b)).$$
Here, the remaining requirement is the same as $sin aleqcos b = sin(pi/2-b)$, which is the same as: $$a+bleqdfrac{pi}{2}.$$
Now let $1>cosleft(frac{a-b}{2}right) = s>cosleft(frac{a+b}{2}right)=tgeqdfrac{1}{sqrt{2}}.$ Then, the equivalent inequality is:
$$frac st>frac 13+frac 12(2s^2-1 - 2t^2+1) = frac 13+s^2-t^2$$
But this one is trivial as:
$$frac 13+s^2-t^2<frac 13+1-frac 12 = frac 56 <1<frac st.$$
answered Dec 6 at 0:37
dezdichado
6,0991929
If you let $x = sin^2 a, y = sin^2 b$ with $0<b<a<pi/2$, then your inequality is equivalent to:
$$sin acos b - sin bcos a +sin^2bsin a - sin^2asin b >dfrac{sin^2 a-sin^2 b}{3}$$ or $$sin(a-b)>(sin a-sin b)left(frac 13+sin asin bright)$$
or $$dfrac{cosleft(frac {a-b}2right)}{cosleft(frac{a+b}{2}right)}>frac{1}{3}+frac 12(cos(a-b)-cos(a+b)).$$
Here, the remaining requirement is the same as $sin aleqcos b = sin(pi/2-b)$, which is the same as: $$a+bleqdfrac{pi}{2}.$$
Now let $1>cosleft(frac{a-b}{2}right) = s>cosleft(frac{a+b}{2}right)=tgeqdfrac{1}{sqrt{2}}.$ Then, the equivalent inequality is:
$$frac st>frac 13+frac 12(2s^2-1 - 2t^2+1) = frac 13+s^2-t^2$$
But this one is trivial as:
$$frac 13+s^2-t^2<frac 13+1-frac 12 = frac 56 <1<frac st.$$
answered Dec 6 at 0:37
dezdichado
6,0991929
answered Dec 6 at 0:37
dezdichado
6,0991929
answered Dec 6 at 0:37
dezdichado
6,0991929
answered Dec 6 at 0:37
dezdichado
6,0991929
6,0991929
add a comment |
add a comment |
up vote
1
down vote
For a traditional kind of proof I use beside $,0<y<x<1,$ and $,x+yleq 1,$ also
first derivation ($:=0$) and second derivation ($neq 0$) of a term to get minimum and maximum.
Equivalent to the given inequation is $enspacedisplaystyle sqrt{x(1-y)}-sqrt{y(1-x)} > xsqrt{y}- ysqrt{x} + frac{x-y}{3}$
and we devide by $,x-y,$ to get $enspacedisplaystyle frac{1}{sqrt{x(1-y)}+sqrt{y(1-x)}} > frac{1}{frac{1}{sqrt{y}}+frac{1}{sqrt{x}}} + frac{1}{3} ,$ .
Using first and second derivation we get
$displaystyle sqrt{x(1-y)}+sqrt{y(1-x)} leq max(sqrt{x(1-y)}+sqrt{y(1-x)}) = 1 enspace$ and
$displaystyle frac{1}{sqrt{y}}+frac{1}{sqrt{x}} geq frac{1}{sqrt{1-x}}+frac{1}{sqrt{x}} geq min(frac{1}{sqrt{1-x}}+frac{1}{sqrt{x}}) = 2sqrt{2}enspace$ so that follows
$displaystyle frac{1}{sqrt{x(1-y)}+sqrt{y(1-x)}} geq 1 > frac{1}{2sqrt{2}} + frac{1}{3} geq frac{1}{frac{1}{sqrt{y}}+frac{1}{sqrt{x}}} + frac{1}{3} enspace$ as wished.
answered Dec 13 at 11:30
user90369
8,193925
add a comment |
up vote
1
down vote
For a traditional kind of proof I use beside $,0<y<x<1,$ and $,x+yleq 1,$ also
first derivation ($:=0$) and second derivation ($neq 0$) of a term to get minimum and maximum.
Equivalent to the given inequation is $enspacedisplaystyle sqrt{x(1-y)}-sqrt{y(1-x)} > xsqrt{y}- ysqrt{x} + frac{x-y}{3}$
and we devide by $,x-y,$ to get $enspacedisplaystyle frac{1}{sqrt{x(1-y)}+sqrt{y(1-x)}} > frac{1}{frac{1}{sqrt{y}}+frac{1}{sqrt{x}}} + frac{1}{3} ,$ .
Using first and second derivation we get
$displaystyle sqrt{x(1-y)}+sqrt{y(1-x)} leq max(sqrt{x(1-y)}+sqrt{y(1-x)}) = 1 enspace$ and
$displaystyle frac{1}{sqrt{y}}+frac{1}{sqrt{x}} geq frac{1}{sqrt{1-x}}+frac{1}{sqrt{x}} geq min(frac{1}{sqrt{1-x}}+frac{1}{sqrt{x}}) = 2sqrt{2}enspace$ so that follows
$displaystyle frac{1}{sqrt{x(1-y)}+sqrt{y(1-x)}} geq 1 > frac{1}{2sqrt{2}} + frac{1}{3} geq frac{1}{frac{1}{sqrt{y}}+frac{1}{sqrt{x}}} + frac{1}{3} enspace$ as wished.
answered Dec 13 at 11:30
user90369
8,193925
add a comment |
up vote
1
down vote
up vote
1
down vote
For a traditional kind of proof I use beside $,0<y<x<1,$ and $,x+yleq 1,$ also
first derivation ($:=0$) and second derivation ($neq 0$) of a term to get minimum and maximum.
Equivalent to the given inequation is $enspacedisplaystyle sqrt{x(1-y)}-sqrt{y(1-x)} > xsqrt{y}- ysqrt{x} + frac{x-y}{3}$
and we devide by $,x-y,$ to get $enspacedisplaystyle frac{1}{sqrt{x(1-y)}+sqrt{y(1-x)}} > frac{1}{frac{1}{sqrt{y}}+frac{1}{sqrt{x}}} + frac{1}{3} ,$ .
Using first and second derivation we get
$displaystyle sqrt{x(1-y)}+sqrt{y(1-x)} leq max(sqrt{x(1-y)}+sqrt{y(1-x)}) = 1 enspace$ and
$displaystyle frac{1}{sqrt{y}}+frac{1}{sqrt{x}} geq frac{1}{sqrt{1-x}}+frac{1}{sqrt{x}} geq min(frac{1}{sqrt{1-x}}+frac{1}{sqrt{x}}) = 2sqrt{2}enspace$ so that follows
$displaystyle frac{1}{sqrt{x(1-y)}+sqrt{y(1-x)}} geq 1 > frac{1}{2sqrt{2}} + frac{1}{3} geq frac{1}{frac{1}{sqrt{y}}+frac{1}{sqrt{x}}} + frac{1}{3} enspace$ as wished.
answered Dec 13 at 11:30
user90369
8,193925
For a traditional kind of proof I use beside $,0<y<x<1,$ and $,x+yleq 1,$ also
first derivation ($:=0$) and second derivation ($neq 0$) of a term to get minimum and maximum.
Equivalent to the given inequation is $enspacedisplaystyle sqrt{x(1-y)}-sqrt{y(1-x)} > xsqrt{y}- ysqrt{x} + frac{x-y}{3}$
and we devide by $,x-y,$ to get $enspacedisplaystyle frac{1}{sqrt{x(1-y)}+sqrt{y(1-x)}} > frac{1}{frac{1}{sqrt{y}}+frac{1}{sqrt{x}}} + frac{1}{3} ,$ .
Using first and second derivation we get
$displaystyle sqrt{x(1-y)}+sqrt{y(1-x)} leq max(sqrt{x(1-y)}+sqrt{y(1-x)}) = 1 enspace$ and
$displaystyle frac{1}{sqrt{y}}+frac{1}{sqrt{x}} geq frac{1}{sqrt{1-x}}+frac{1}{sqrt{x}} geq min(frac{1}{sqrt{1-x}}+frac{1}{sqrt{x}}) = 2sqrt{2}enspace$ so that follows
$displaystyle frac{1}{sqrt{x(1-y)}+sqrt{y(1-x)}} geq 1 > frac{1}{2sqrt{2}} + frac{1}{3} geq frac{1}{frac{1}{sqrt{y}}+frac{1}{sqrt{x}}} + frac{1}{3} enspace$ as wished.
answered Dec 13 at 11:30
user90369
8,193925
answered Dec 13 at 11:30
user90369
8,193925
answered Dec 13 at 11:30
user90369
8,193925
answered Dec 13 at 11:30
user90369
8,193925
8,193925
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3027476%2fshow-sqrtx-1-y-sqrty-1-x-y-sqrtx-x-sqrty-fracx-y3%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
