Problem wih hyperbolic function
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1
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Can we use hyperbolic function to solve the following problems ?
If $(sqrt {{y^2-x^3}} - x)(sqrt {{x^2} + y^3} - y) =y^3$ , prove that $x+ y = 0$
If $(sqrt {{x^2+y^4}} - x)(sqrt {{y^2} + x^4} - y) le x^2 y^2$ , prove that $x + y ge 0$
inequality exponential-function hyperbolic-equations
asked Nov 9 at 4:54
Trần Văn Lâm
412
|
show 6 more comments
up vote
1
down vote
favorite
Can we use hyperbolic function to solve the following problems ?
If $(sqrt {{y^2-x^3}} - x)(sqrt {{x^2} + y^3} - y) =y^3$ , prove that $x+ y = 0$
If $(sqrt {{x^2+y^4}} - x)(sqrt {{y^2} + x^4} - y) le x^2 y^2$ , prove that $x + y ge 0$
inequality exponential-function hyperbolic-equations
asked Nov 9 at 4:54
Trần Văn Lâm
412
Do you want see a solution without hyperbolic functions?
– Michael Rozenberg
Nov 9 at 5:32
Yes, of course! Michael Rozenberg, please post your full solutions.
– Trần Văn Lâm
Nov 9 at 9:10
1
You need to show us your trying, otherwise this topic will be closed.
– Michael Rozenberg
Nov 9 at 12:08
Sir, I thought $x=sinh t, ycosh t$. And nothing more!
– Trần Văn Lâm
Nov 10 at 0:08
Sir Michael Rozenberg, please! I can't wait because I want to see your solution.
– Trần Văn Lâm
Nov 10 at 13:05
|
show 6 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Can we use hyperbolic function to solve the following problems ?
If $(sqrt {{y^2-x^3}} - x)(sqrt {{x^2} + y^3} - y) =y^3$ , prove that $x+ y = 0$
If $(sqrt {{x^2+y^4}} - x)(sqrt {{y^2} + x^4} - y) le x^2 y^2$ , prove that $x + y ge 0$
inequality exponential-function hyperbolic-equations
asked Nov 9 at 4:54
Trần Văn Lâm
412
Can we use hyperbolic function to solve the following problems ?
If $(sqrt {{y^2-x^3}} - x)(sqrt {{x^2} + y^3} - y) =y^3$ , prove that $x+ y = 0$
If $(sqrt {{x^2+y^4}} - x)(sqrt {{y^2} + x^4} - y) le x^2 y^2$ , prove that $x + y ge 0$
inequality exponential-function hyperbolic-equations
inequality exponential-function hyperbolic-equations
asked Nov 9 at 4:54
Trần Văn Lâm
412
asked Nov 9 at 4:54
Trần Văn Lâm
412
asked Nov 9 at 4:54
Trần Văn Lâm
412
asked Nov 9 at 4:54
Trần Văn Lâm
412
asked Nov 9 at 4:54
Trần Văn Lâm
412
412
Do you want see a solution without hyperbolic functions?
– Michael Rozenberg
Nov 9 at 5:32
Yes, of course! Michael Rozenberg, please post your full solutions.
– Trần Văn Lâm
Nov 9 at 9:10
1
You need to show us your trying, otherwise this topic will be closed.
– Michael Rozenberg
Nov 9 at 12:08
Sir, I thought $x=sinh t, ycosh t$. And nothing more!
– Trần Văn Lâm
Nov 10 at 0:08
Sir Michael Rozenberg, please! I can't wait because I want to see your solution.
– Trần Văn Lâm
Nov 10 at 13:05
|
show 6 more comments
Do you want see a solution without hyperbolic functions?
– Michael Rozenberg
Nov 9 at 5:32
Yes, of course! Michael Rozenberg, please post your full solutions.
– Trần Văn Lâm
Nov 9 at 9:10
1
You need to show us your trying, otherwise this topic will be closed.
– Michael Rozenberg
Nov 9 at 12:08
Sir, I thought $x=sinh t, ycosh t$. And nothing more!
– Trần Văn Lâm
Nov 10 at 0:08
Sir Michael Rozenberg, please! I can't wait because I want to see your solution.
– Trần Văn Lâm
Nov 10 at 13:05
Do you want see a solution without hyperbolic functions?
– Michael Rozenberg
Nov 9 at 5:32
Do you want see a solution without hyperbolic functions?
– Michael Rozenberg
Nov 9 at 5:32
Yes, of course! Michael Rozenberg, please post your full solutions.
– Trần Văn Lâm
Nov 9 at 9:10
Yes, of course! Michael Rozenberg, please post your full solutions.
– Trần Văn Lâm
Nov 9 at 9:10
1
1
You need to show us your trying, otherwise this topic will be closed.
– Michael Rozenberg
Nov 9 at 12:08
You need to show us your trying, otherwise this topic will be closed.
– Michael Rozenberg
Nov 9 at 12:08
Sir, I thought $x=sinh t, ycosh t$. And nothing more!
– Trần Văn Lâm
Nov 10 at 0:08
Sir, I thought $x=sinh t, ycosh t$. And nothing more!
– Trần Văn Lâm
Nov 10 at 0:08
Sir Michael Rozenberg, please! I can't wait because I want to see your solution.
– Trần Văn Lâm
Nov 10 at 13:05
Sir Michael Rozenberg, please! I can't wait because I want to see your solution.
– Trần Văn Lâm
Nov 10 at 13:05
|
show 6 more comments
1 Answer
1
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oldest
votes
up vote
1
down vote
By C-S $$x^2y^2geqsqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xygeq$$
$$geq |xy|+x^2y^2-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xy,$$
which gives
$$xsqrt{y^2+x^4}+ysqrt{x^2+y^4}geq|xy|+xy.$$
Now, if $xygeq0$ so since
$$xsqrt{y^2+x^4}+ysqrt{x^2+y^4}geq0,$$ we obtain $xgeq0$, $ygeq0$ and from here $x+ygeq0.$
Let $xy<0$, $x>0$ and $y<0$.
Thus, $$xsqrt{y^2+x^4}geq-ysqrt{x^2+y^4}$$ or
$$x^2y^2+x^6geq x^2y^2+y^6$$ or
$$x^2geq y^2,$$ which gives $x+ygeq0$ because $x-y>0.$
answered Nov 23 at 8:48
Michael Rozenberg
94.4k1588183
The first part, $x^2y^2geqsqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xy$ is what you WANT to establish. The second part, $sqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xygeq |xy|+x^2y^2-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}$ doesn't hold true in general, e.g. $(x,y)=(1,−1)$ doesn't hold. So I wonder how the proof goes.
– Andreas
2 days ago
@Andreas I work with the given. Your example $(1,-1)$ is not valid. See please better the given.
– Michael Rozenberg
2 days ago
The question doen't say anything about the signs or ranges of $x$ and $y$.
– Andreas
2 days ago
1
Ok I got it. You are of course right that on the condition side, we can make that replacement.
– Andreas
2 days ago
1
Ok I followed the further steps in your answer and it looks real fine. Congrats!
– Andreas
2 days ago
|
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
By C-S $$x^2y^2geqsqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xygeq$$
$$geq |xy|+x^2y^2-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xy,$$
which gives
$$xsqrt{y^2+x^4}+ysqrt{x^2+y^4}geq|xy|+xy.$$
Now, if $xygeq0$ so since
$$xsqrt{y^2+x^4}+ysqrt{x^2+y^4}geq0,$$ we obtain $xgeq0$, $ygeq0$ and from here $x+ygeq0.$
Let $xy<0$, $x>0$ and $y<0$.
Thus, $$xsqrt{y^2+x^4}geq-ysqrt{x^2+y^4}$$ or
$$x^2y^2+x^6geq x^2y^2+y^6$$ or
$$x^2geq y^2,$$ which gives $x+ygeq0$ because $x-y>0.$
answered Nov 23 at 8:48
Michael Rozenberg
94.4k1588183
The first part, $x^2y^2geqsqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xy$ is what you WANT to establish. The second part, $sqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xygeq |xy|+x^2y^2-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}$ doesn't hold true in general, e.g. $(x,y)=(1,−1)$ doesn't hold. So I wonder how the proof goes.
– Andreas
2 days ago
@Andreas I work with the given. Your example $(1,-1)$ is not valid. See please better the given.
– Michael Rozenberg
2 days ago
The question doen't say anything about the signs or ranges of $x$ and $y$.
– Andreas
2 days ago
1
Ok I got it. You are of course right that on the condition side, we can make that replacement.
– Andreas
2 days ago
1
Ok I followed the further steps in your answer and it looks real fine. Congrats!
– Andreas
2 days ago
|
show 3 more comments
up vote
1
down vote
By C-S $$x^2y^2geqsqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xygeq$$
$$geq |xy|+x^2y^2-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xy,$$
which gives
$$xsqrt{y^2+x^4}+ysqrt{x^2+y^4}geq|xy|+xy.$$
Now, if $xygeq0$ so since
$$xsqrt{y^2+x^4}+ysqrt{x^2+y^4}geq0,$$ we obtain $xgeq0$, $ygeq0$ and from here $x+ygeq0.$
Let $xy<0$, $x>0$ and $y<0$.
Thus, $$xsqrt{y^2+x^4}geq-ysqrt{x^2+y^4}$$ or
$$x^2y^2+x^6geq x^2y^2+y^6$$ or
$$x^2geq y^2,$$ which gives $x+ygeq0$ because $x-y>0.$
answered Nov 23 at 8:48
Michael Rozenberg
94.4k1588183
The first part, $x^2y^2geqsqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xy$ is what you WANT to establish. The second part, $sqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xygeq |xy|+x^2y^2-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}$ doesn't hold true in general, e.g. $(x,y)=(1,−1)$ doesn't hold. So I wonder how the proof goes.
– Andreas
2 days ago
@Andreas I work with the given. Your example $(1,-1)$ is not valid. See please better the given.
– Michael Rozenberg
2 days ago
The question doen't say anything about the signs or ranges of $x$ and $y$.
– Andreas
2 days ago
1
Ok I got it. You are of course right that on the condition side, we can make that replacement.
– Andreas
2 days ago
1
Ok I followed the further steps in your answer and it looks real fine. Congrats!
– Andreas
2 days ago
|
show 3 more comments
up vote
1
down vote
up vote
1
down vote
By C-S $$x^2y^2geqsqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xygeq$$
$$geq |xy|+x^2y^2-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xy,$$
which gives
$$xsqrt{y^2+x^4}+ysqrt{x^2+y^4}geq|xy|+xy.$$
Now, if $xygeq0$ so since
$$xsqrt{y^2+x^4}+ysqrt{x^2+y^4}geq0,$$ we obtain $xgeq0$, $ygeq0$ and from here $x+ygeq0.$
Let $xy<0$, $x>0$ and $y<0$.
Thus, $$xsqrt{y^2+x^4}geq-ysqrt{x^2+y^4}$$ or
$$x^2y^2+x^6geq x^2y^2+y^6$$ or
$$x^2geq y^2,$$ which gives $x+ygeq0$ because $x-y>0.$
answered Nov 23 at 8:48
Michael Rozenberg
94.4k1588183
By C-S $$x^2y^2geqsqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xygeq$$
$$geq |xy|+x^2y^2-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xy,$$
which gives
$$xsqrt{y^2+x^4}+ysqrt{x^2+y^4}geq|xy|+xy.$$
Now, if $xygeq0$ so since
$$xsqrt{y^2+x^4}+ysqrt{x^2+y^4}geq0,$$ we obtain $xgeq0$, $ygeq0$ and from here $x+ygeq0.$
Let $xy<0$, $x>0$ and $y<0$.
Thus, $$xsqrt{y^2+x^4}geq-ysqrt{x^2+y^4}$$ or
$$x^2y^2+x^6geq x^2y^2+y^6$$ or
$$x^2geq y^2,$$ which gives $x+ygeq0$ because $x-y>0.$
answered Nov 23 at 8:48
Michael Rozenberg
94.4k1588183
edited 2 days ago
answered Nov 23 at 8:48
Michael Rozenberg
94.4k1588183
answered Nov 23 at 8:48
Michael Rozenberg
94.4k1588183
answered Nov 23 at 8:48
Michael Rozenberg
94.4k1588183
94.4k1588183
The first part, $x^2y^2geqsqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xy$ is what you WANT to establish. The second part, $sqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xygeq |xy|+x^2y^2-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}$ doesn't hold true in general, e.g. $(x,y)=(1,−1)$ doesn't hold. So I wonder how the proof goes.
– Andreas
2 days ago
@Andreas I work with the given. Your example $(1,-1)$ is not valid. See please better the given.
– Michael Rozenberg
2 days ago
The question doen't say anything about the signs or ranges of $x$ and $y$.
– Andreas
2 days ago
1
Ok I got it. You are of course right that on the condition side, we can make that replacement.
– Andreas
2 days ago
1
Ok I followed the further steps in your answer and it looks real fine. Congrats!
– Andreas
2 days ago
|
show 3 more comments
The first part, $x^2y^2geqsqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xy$ is what you WANT to establish. The second part, $sqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xygeq |xy|+x^2y^2-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}$ doesn't hold true in general, e.g. $(x,y)=(1,−1)$ doesn't hold. So I wonder how the proof goes.
– Andreas
2 days ago
@Andreas I work with the given. Your example $(1,-1)$ is not valid. See please better the given.
– Michael Rozenberg
2 days ago
The question doen't say anything about the signs or ranges of $x$ and $y$.
– Andreas
2 days ago
1
Ok I got it. You are of course right that on the condition side, we can make that replacement.
– Andreas
2 days ago
1
Ok I followed the further steps in your answer and it looks real fine. Congrats!
– Andreas
2 days ago
The first part, $x^2y^2geqsqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xy$ is what you WANT to establish. The second part, $sqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xygeq |xy|+x^2y^2-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}$ doesn't hold true in general, e.g. $(x,y)=(1,−1)$ doesn't hold. So I wonder how the proof goes.
– Andreas
2 days ago
The first part, $x^2y^2geqsqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xy$ is what you WANT to establish. The second part, $sqrt{(x^2+y^4)(y^2+x^4)}-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}+xygeq |xy|+x^2y^2-xsqrt{y^2+x^4}-ysqrt{x^2+y^4}$ doesn't hold true in general, e.g. $(x,y)=(1,−1)$ doesn't hold. So I wonder how the proof goes.
– Andreas
2 days ago
@Andreas I work with the given. Your example $(1,-1)$ is not valid. See please better the given.
– Michael Rozenberg
2 days ago
@Andreas I work with the given. Your example $(1,-1)$ is not valid. See please better the given.
– Michael Rozenberg
2 days ago
The question doen't say anything about the signs or ranges of $x$ and $y$.
– Andreas
2 days ago
The question doen't say anything about the signs or ranges of $x$ and $y$.
– Andreas
2 days ago
1
1
Ok I got it. You are of course right that on the condition side, we can make that replacement.
– Andreas
2 days ago
Ok I got it. You are of course right that on the condition side, we can make that replacement.
– Andreas
2 days ago
1
1
Ok I followed the further steps in your answer and it looks real fine. Congrats!
– Andreas
2 days ago
Ok I followed the further steps in your answer and it looks real fine. Congrats!
– Andreas
2 days ago
|
show 3 more comments
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Do you want see a solution without hyperbolic functions?
– Michael Rozenberg
Nov 9 at 5:32
Yes, of course! Michael Rozenberg, please post your full solutions.
– Trần Văn Lâm
Nov 9 at 9:10
1
You need to show us your trying, otherwise this topic will be closed.
– Michael Rozenberg
Nov 9 at 12:08
Sir, I thought $x=sinh t, ycosh t$. And nothing more!
– Trần Văn Lâm
Nov 10 at 0:08
Sir Michael Rozenberg, please! I can't wait because I want to see your solution.
– Trần Văn Lâm
Nov 10 at 13:05