show $sum_{cyc}(1-x)^2ge sum_{cyc}frac{z^2(1-x^2)(1-y^2)}{(xy+z)^2}.$
$begingroup$
Let $x,y,z>0$. Show that
$$sum_{cyc}(1-x)^2ge sum_{cyc}dfrac{z^2(1-x^2)(1-y^2)}{(xy+z)^2}.$$
Here's what I have done. The expression
$$sum_{cyc}(1-x)left((1-x)-dfrac{z^2(1+x)(1-y^2)}{(xy+z)^2}right)ge 0$$
or
$$sum_{cyc}(1-x)cdotdfrac{x^2y^2+y^2z^2+xy^2z^2+2xyz-2xz^2-2x^2yz-x^3y^2}{(xy+z)^2}ge 0.$$
But this way does not help for the starting inequality.
inequality
edited Jul 5 '17 at 11:11
Arjang
5,61562363
asked Jun 21 '17 at 0:56
communnitescommunnites
1,1931534
$endgroup$
add a comment |
$begingroup$
Let $x,y,z>0$. Show that
$$sum_{cyc}(1-x)^2ge sum_{cyc}dfrac{z^2(1-x^2)(1-y^2)}{(xy+z)^2}.$$
Here's what I have done. The expression
$$sum_{cyc}(1-x)left((1-x)-dfrac{z^2(1+x)(1-y^2)}{(xy+z)^2}right)ge 0$$
or
$$sum_{cyc}(1-x)cdotdfrac{x^2y^2+y^2z^2+xy^2z^2+2xyz-2xz^2-2x^2yz-x^3y^2}{(xy+z)^2}ge 0.$$
But this way does not help for the starting inequality.
inequality
edited Jul 5 '17 at 11:11
Arjang
5,61562363
asked Jun 21 '17 at 0:56
communnitescommunnites
1,1931534
$endgroup$
$begingroup$
@Rozenberg Is inequality true for all nonnegatives? Or is there counter-exs ?
$endgroup$
– Merdanov
Jun 22 '17 at 12:00
$begingroup$
maybe let $y=x+p, z=x+p+q.p,qge 0$ is usefull?
$endgroup$
– communnites
Jun 29 '17 at 4:54
$begingroup$
I have a proof for $x+y+zgeqfrac{3}{2}.$
$endgroup$
– Michael Rozenberg
Dec 21 '18 at 13:01
$begingroup$
Hint became the answer.
$endgroup$
– Yuri Negometyanov
Dec 27 '18 at 14:49
add a comment |
$begingroup$
Let $x,y,z>0$. Show that
$$sum_{cyc}(1-x)^2ge sum_{cyc}dfrac{z^2(1-x^2)(1-y^2)}{(xy+z)^2}.$$
Here's what I have done. The expression
$$sum_{cyc}(1-x)left((1-x)-dfrac{z^2(1+x)(1-y^2)}{(xy+z)^2}right)ge 0$$
or
$$sum_{cyc}(1-x)cdotdfrac{x^2y^2+y^2z^2+xy^2z^2+2xyz-2xz^2-2x^2yz-x^3y^2}{(xy+z)^2}ge 0.$$
But this way does not help for the starting inequality.
inequality
edited Jul 5 '17 at 11:11
Arjang
5,61562363
asked Jun 21 '17 at 0:56
communnitescommunnites
1,1931534
$endgroup$
Let $x,y,z>0$. Show that
$$sum_{cyc}(1-x)^2ge sum_{cyc}dfrac{z^2(1-x^2)(1-y^2)}{(xy+z)^2}.$$
Here's what I have done. The expression
$$sum_{cyc}(1-x)left((1-x)-dfrac{z^2(1+x)(1-y^2)}{(xy+z)^2}right)ge 0$$
or
$$sum_{cyc}(1-x)cdotdfrac{x^2y^2+y^2z^2+xy^2z^2+2xyz-2xz^2-2x^2yz-x^3y^2}{(xy+z)^2}ge 0.$$
But this way does not help for the starting inequality.
inequality
inequality
edited Jul 5 '17 at 11:11
Arjang
5,61562363
asked Jun 21 '17 at 0:56
communnitescommunnites
1,1931534
edited Jul 5 '17 at 11:11
Arjang
5,61562363
asked Jun 21 '17 at 0:56
communnitescommunnites
1,1931534
edited Jul 5 '17 at 11:11
Arjang
5,61562363
edited Jul 5 '17 at 11:11
Arjang
5,61562363
edited Jul 5 '17 at 11:11
Arjang
5,61562363
5,61562363
asked Jun 21 '17 at 0:56
communnitescommunnites
1,1931534
asked Jun 21 '17 at 0:56
communnitescommunnites
1,1931534
asked Jun 21 '17 at 0:56
communnitescommunnites
1,1931534
1,1931534
$begingroup$
@Rozenberg Is inequality true for all nonnegatives? Or is there counter-exs ?
$endgroup$
– Merdanov
Jun 22 '17 at 12:00
$begingroup$
maybe let $y=x+p, z=x+p+q.p,qge 0$ is usefull?
$endgroup$
– communnites
Jun 29 '17 at 4:54
$begingroup$
I have a proof for $x+y+zgeqfrac{3}{2}.$
$endgroup$
– Michael Rozenberg
Dec 21 '18 at 13:01
$begingroup$
Hint became the answer.
$endgroup$
– Yuri Negometyanov
Dec 27 '18 at 14:49
add a comment |
$begingroup$
@Rozenberg Is inequality true for all nonnegatives? Or is there counter-exs ?
$endgroup$
– Merdanov
Jun 22 '17 at 12:00
$begingroup$
maybe let $y=x+p, z=x+p+q.p,qge 0$ is usefull?
$endgroup$
– communnites
Jun 29 '17 at 4:54
$begingroup$
I have a proof for $x+y+zgeqfrac{3}{2}.$
$endgroup$
– Michael Rozenberg
Dec 21 '18 at 13:01
$begingroup$
Hint became the answer.
$endgroup$
– Yuri Negometyanov
Dec 27 '18 at 14:49
$begingroup$
@Rozenberg Is inequality true for all nonnegatives? Or is there counter-exs ?
$endgroup$
– Merdanov
Jun 22 '17 at 12:00
$begingroup$
@Rozenberg Is inequality true for all nonnegatives? Or is there counter-exs ?
$endgroup$
– Merdanov
Jun 22 '17 at 12:00
$begingroup$
maybe let $y=x+p, z=x+p+q.p,qge 0$ is usefull?
$endgroup$
– communnites
Jun 29 '17 at 4:54
$begingroup$
maybe let $y=x+p, z=x+p+q.p,qge 0$ is usefull?
$endgroup$
– communnites
Jun 29 '17 at 4:54
$begingroup$
I have a proof for $x+y+zgeqfrac{3}{2}.$
$endgroup$
– Michael Rozenberg
Dec 21 '18 at 13:01
$begingroup$
I have a proof for $x+y+zgeqfrac{3}{2}.$
$endgroup$
– Michael Rozenberg
Dec 21 '18 at 13:01
$begingroup$
Hint became the answer.
$endgroup$
– Yuri Negometyanov
Dec 27 '18 at 14:49
$begingroup$
Hint became the answer.
$endgroup$
– Yuri Negometyanov
Dec 27 '18 at 14:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The idea to prove this inequality is to use the well-know Ky-Fan inequality :
Before that we make the following substitution:
$x=frac{1}{a}$
$y=frac{1}{b}$
$z=frac{1}{c}$
We get the following inequality :
$$sum_{cyc}frac{(1-a)^2}{a^2}geq sum_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$$
Now we have the Ky-fan inequality wich state that for $a,b,c$ belongs to $[frac{1}{4};frac{1}{2}]$:
$$frac{(1-a)^2}{a^2}geqfrac{b^2c^2}{(1-b)^2(1-c)^2}Big(frac{3-a-b-c}{a+b+c}Big)^6
$$
So we get this refinement of the starting inequality if we sum each term of the previous inequality:
$$sum_{cyc}frac{(1-a)^2}{a^2}geq sum_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}Big(frac{3-a-b-c}{a+b+c}Big)^6 geqsum_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$$
So we prove the inequality for $a,b,c$ belongs to $[frac{1}{4};frac{1}{2}]$
To prove the remainder of the theorem you just have to play with a variable $u$ and remark that we always have for all positive $x,y,z$:
$$sum_{cyc}frac{(1-x)^2}{x^2}geq usum_{cyc}frac{(1-a)^2}{a^2}geq usum_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}Big(frac{3-a-b-c}{a+b+c}Big)^6\ geq usum_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}geq sum_{cyc}frac{(1-y^2)(1-z^2)}{(yz+x)^2}$$
We make the following substitution on the first inequality :
$x=cos^2 s$;
$y=cos^2 v$;
$z=cos^2 w$
And
$a=cos^2 p$;
$b=cos^2 r$;
$c=cos^2 q$
With $p,q,r$ belonging to $[arccos (frac{1}{sqrt{2}});arccos(frac{1}{2})]$ :
We get this :
$$sum_{cyc}tan^4 sgeq usum_{cyc}tan^4 p$$
we put :
$$u=frac{sum_{cyc}tan^2 stan^2 v}{sum_{cyc}tan^4 p}$$
Edit for Martin R :
We have build $u$ exactly to obtain the first inequality :
So we have :
$$usum_{cyc}tan^4 p=sum_{cyc}tan^2 stan^2 v$$
And
$$sum_{cyc}tan^4 sgeqsum_{cyc}tan^2 stan^2 v$$
And to conclude it's just Am-Gm or $a^2+b^2geq 2ab$
Now we need to prove that $u$ satisfy the last inequality :
With the previous substitution and combine with the well-know formula $(cos a^2+sin a^2)^2=1$ we find:
$$frac{sum_{cyc}tan^2 stan v^2}{sum_{cyc}tan^4 p}sum_{cyc}frac{(tan^4 p+2tan^2 p)((tan^4 r+2tan^2 r)}{Big(1+frac{cos^2 q}{cos^2 rcos^2 p}Big)^2}geqsum_{cyc}frac{(tan^4 s+2tan^2 s)((tan^4 v+2tan^2 v)}{Big(1+frac{cos^2 w}{cos^2 vcos^2 s}Big)^2}$$
$endgroup$
1
$begingroup$
Why $sumlimits_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}left(frac{3-a-b-c}{a+b+c}right)^6 geqsumlimits_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$ is true?
$endgroup$
– Michael Rozenberg
Jul 3 '17 at 8:15
$begingroup$
See here.It works because there is no order on $a,b,c$.
$endgroup$
– user448747
Jul 3 '17 at 12:55
$begingroup$
Explain please your play with a variable $u$ for the rest cases.
$endgroup$
– Michael Rozenberg
Jul 3 '17 at 13:16
$begingroup$
If you use logarithmic majorization you will find a particular value for $u$ ;$u=frac{(1-x)^2a^2}{x^2(1-a)^2}$
$endgroup$
– user448747
Jul 3 '17 at 15:06
1
$begingroup$
Can someone, which upper voted this post, explain me why this solution is true? I think it's nothing.
$endgroup$
– Michael Rozenberg
Jul 6 '17 at 5:30
|
show 5 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The idea to prove this inequality is to use the well-know Ky-Fan inequality :
Before that we make the following substitution:
$x=frac{1}{a}$
$y=frac{1}{b}$
$z=frac{1}{c}$
We get the following inequality :
$$sum_{cyc}frac{(1-a)^2}{a^2}geq sum_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$$
Now we have the Ky-fan inequality wich state that for $a,b,c$ belongs to $[frac{1}{4};frac{1}{2}]$:
$$frac{(1-a)^2}{a^2}geqfrac{b^2c^2}{(1-b)^2(1-c)^2}Big(frac{3-a-b-c}{a+b+c}Big)^6
$$
So we get this refinement of the starting inequality if we sum each term of the previous inequality:
$$sum_{cyc}frac{(1-a)^2}{a^2}geq sum_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}Big(frac{3-a-b-c}{a+b+c}Big)^6 geqsum_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$$
So we prove the inequality for $a,b,c$ belongs to $[frac{1}{4};frac{1}{2}]$
To prove the remainder of the theorem you just have to play with a variable $u$ and remark that we always have for all positive $x,y,z$:
$$sum_{cyc}frac{(1-x)^2}{x^2}geq usum_{cyc}frac{(1-a)^2}{a^2}geq usum_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}Big(frac{3-a-b-c}{a+b+c}Big)^6\ geq usum_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}geq sum_{cyc}frac{(1-y^2)(1-z^2)}{(yz+x)^2}$$
We make the following substitution on the first inequality :
$x=cos^2 s$;
$y=cos^2 v$;
$z=cos^2 w$
And
$a=cos^2 p$;
$b=cos^2 r$;
$c=cos^2 q$
With $p,q,r$ belonging to $[arccos (frac{1}{sqrt{2}});arccos(frac{1}{2})]$ :
We get this :
$$sum_{cyc}tan^4 sgeq usum_{cyc}tan^4 p$$
we put :
$$u=frac{sum_{cyc}tan^2 stan^2 v}{sum_{cyc}tan^4 p}$$
Edit for Martin R :
We have build $u$ exactly to obtain the first inequality :
So we have :
$$usum_{cyc}tan^4 p=sum_{cyc}tan^2 stan^2 v$$
And
$$sum_{cyc}tan^4 sgeqsum_{cyc}tan^2 stan^2 v$$
And to conclude it's just Am-Gm or $a^2+b^2geq 2ab$
Now we need to prove that $u$ satisfy the last inequality :
With the previous substitution and combine with the well-know formula $(cos a^2+sin a^2)^2=1$ we find:
$$frac{sum_{cyc}tan^2 stan v^2}{sum_{cyc}tan^4 p}sum_{cyc}frac{(tan^4 p+2tan^2 p)((tan^4 r+2tan^2 r)}{Big(1+frac{cos^2 q}{cos^2 rcos^2 p}Big)^2}geqsum_{cyc}frac{(tan^4 s+2tan^2 s)((tan^4 v+2tan^2 v)}{Big(1+frac{cos^2 w}{cos^2 vcos^2 s}Big)^2}$$
$endgroup$
1
$begingroup$
Why $sumlimits_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}left(frac{3-a-b-c}{a+b+c}right)^6 geqsumlimits_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$ is true?
$endgroup$
– Michael Rozenberg
Jul 3 '17 at 8:15
$begingroup$
See here.It works because there is no order on $a,b,c$.
$endgroup$
– user448747
Jul 3 '17 at 12:55
$begingroup$
Explain please your play with a variable $u$ for the rest cases.
$endgroup$
– Michael Rozenberg
Jul 3 '17 at 13:16
$begingroup$
If you use logarithmic majorization you will find a particular value for $u$ ;$u=frac{(1-x)^2a^2}{x^2(1-a)^2}$
$endgroup$
– user448747
Jul 3 '17 at 15:06
1
$begingroup$
Can someone, which upper voted this post, explain me why this solution is true? I think it's nothing.
$endgroup$
– Michael Rozenberg
Jul 6 '17 at 5:30
|
show 5 more comments
$begingroup$
The idea to prove this inequality is to use the well-know Ky-Fan inequality :
Before that we make the following substitution:
$x=frac{1}{a}$
$y=frac{1}{b}$
$z=frac{1}{c}$
We get the following inequality :
$$sum_{cyc}frac{(1-a)^2}{a^2}geq sum_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$$
Now we have the Ky-fan inequality wich state that for $a,b,c$ belongs to $[frac{1}{4};frac{1}{2}]$:
$$frac{(1-a)^2}{a^2}geqfrac{b^2c^2}{(1-b)^2(1-c)^2}Big(frac{3-a-b-c}{a+b+c}Big)^6
$$
So we get this refinement of the starting inequality if we sum each term of the previous inequality:
$$sum_{cyc}frac{(1-a)^2}{a^2}geq sum_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}Big(frac{3-a-b-c}{a+b+c}Big)^6 geqsum_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$$
So we prove the inequality for $a,b,c$ belongs to $[frac{1}{4};frac{1}{2}]$
To prove the remainder of the theorem you just have to play with a variable $u$ and remark that we always have for all positive $x,y,z$:
$$sum_{cyc}frac{(1-x)^2}{x^2}geq usum_{cyc}frac{(1-a)^2}{a^2}geq usum_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}Big(frac{3-a-b-c}{a+b+c}Big)^6\ geq usum_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}geq sum_{cyc}frac{(1-y^2)(1-z^2)}{(yz+x)^2}$$
We make the following substitution on the first inequality :
$x=cos^2 s$;
$y=cos^2 v$;
$z=cos^2 w$
And
$a=cos^2 p$;
$b=cos^2 r$;
$c=cos^2 q$
With $p,q,r$ belonging to $[arccos (frac{1}{sqrt{2}});arccos(frac{1}{2})]$ :
We get this :
$$sum_{cyc}tan^4 sgeq usum_{cyc}tan^4 p$$
we put :
$$u=frac{sum_{cyc}tan^2 stan^2 v}{sum_{cyc}tan^4 p}$$
Edit for Martin R :
We have build $u$ exactly to obtain the first inequality :
So we have :
$$usum_{cyc}tan^4 p=sum_{cyc}tan^2 stan^2 v$$
And
$$sum_{cyc}tan^4 sgeqsum_{cyc}tan^2 stan^2 v$$
And to conclude it's just Am-Gm or $a^2+b^2geq 2ab$
Now we need to prove that $u$ satisfy the last inequality :
With the previous substitution and combine with the well-know formula $(cos a^2+sin a^2)^2=1$ we find:
$$frac{sum_{cyc}tan^2 stan v^2}{sum_{cyc}tan^4 p}sum_{cyc}frac{(tan^4 p+2tan^2 p)((tan^4 r+2tan^2 r)}{Big(1+frac{cos^2 q}{cos^2 rcos^2 p}Big)^2}geqsum_{cyc}frac{(tan^4 s+2tan^2 s)((tan^4 v+2tan^2 v)}{Big(1+frac{cos^2 w}{cos^2 vcos^2 s}Big)^2}$$
$endgroup$
1
$begingroup$
Why $sumlimits_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}left(frac{3-a-b-c}{a+b+c}right)^6 geqsumlimits_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$ is true?
$endgroup$
– Michael Rozenberg
Jul 3 '17 at 8:15
$begingroup$
See here.It works because there is no order on $a,b,c$.
$endgroup$
– user448747
Jul 3 '17 at 12:55
$begingroup$
Explain please your play with a variable $u$ for the rest cases.
$endgroup$
– Michael Rozenberg
Jul 3 '17 at 13:16
$begingroup$
If you use logarithmic majorization you will find a particular value for $u$ ;$u=frac{(1-x)^2a^2}{x^2(1-a)^2}$
$endgroup$
– user448747
Jul 3 '17 at 15:06
1
$begingroup$
Can someone, which upper voted this post, explain me why this solution is true? I think it's nothing.
$endgroup$
– Michael Rozenberg
Jul 6 '17 at 5:30
|
show 5 more comments
$begingroup$
The idea to prove this inequality is to use the well-know Ky-Fan inequality :
Before that we make the following substitution:
$x=frac{1}{a}$
$y=frac{1}{b}$
$z=frac{1}{c}$
We get the following inequality :
$$sum_{cyc}frac{(1-a)^2}{a^2}geq sum_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$$
Now we have the Ky-fan inequality wich state that for $a,b,c$ belongs to $[frac{1}{4};frac{1}{2}]$:
$$frac{(1-a)^2}{a^2}geqfrac{b^2c^2}{(1-b)^2(1-c)^2}Big(frac{3-a-b-c}{a+b+c}Big)^6
$$
So we get this refinement of the starting inequality if we sum each term of the previous inequality:
$$sum_{cyc}frac{(1-a)^2}{a^2}geq sum_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}Big(frac{3-a-b-c}{a+b+c}Big)^6 geqsum_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$$
So we prove the inequality for $a,b,c$ belongs to $[frac{1}{4};frac{1}{2}]$
To prove the remainder of the theorem you just have to play with a variable $u$ and remark that we always have for all positive $x,y,z$:
$$sum_{cyc}frac{(1-x)^2}{x^2}geq usum_{cyc}frac{(1-a)^2}{a^2}geq usum_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}Big(frac{3-a-b-c}{a+b+c}Big)^6\ geq usum_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}geq sum_{cyc}frac{(1-y^2)(1-z^2)}{(yz+x)^2}$$
We make the following substitution on the first inequality :
$x=cos^2 s$;
$y=cos^2 v$;
$z=cos^2 w$
And
$a=cos^2 p$;
$b=cos^2 r$;
$c=cos^2 q$
With $p,q,r$ belonging to $[arccos (frac{1}{sqrt{2}});arccos(frac{1}{2})]$ :
We get this :
$$sum_{cyc}tan^4 sgeq usum_{cyc}tan^4 p$$
we put :
$$u=frac{sum_{cyc}tan^2 stan^2 v}{sum_{cyc}tan^4 p}$$
Edit for Martin R :
We have build $u$ exactly to obtain the first inequality :
So we have :
$$usum_{cyc}tan^4 p=sum_{cyc}tan^2 stan^2 v$$
And
$$sum_{cyc}tan^4 sgeqsum_{cyc}tan^2 stan^2 v$$
And to conclude it's just Am-Gm or $a^2+b^2geq 2ab$
Now we need to prove that $u$ satisfy the last inequality :
With the previous substitution and combine with the well-know formula $(cos a^2+sin a^2)^2=1$ we find:
$$frac{sum_{cyc}tan^2 stan v^2}{sum_{cyc}tan^4 p}sum_{cyc}frac{(tan^4 p+2tan^2 p)((tan^4 r+2tan^2 r)}{Big(1+frac{cos^2 q}{cos^2 rcos^2 p}Big)^2}geqsum_{cyc}frac{(tan^4 s+2tan^2 s)((tan^4 v+2tan^2 v)}{Big(1+frac{cos^2 w}{cos^2 vcos^2 s}Big)^2}$$
$endgroup$
The idea to prove this inequality is to use the well-know Ky-Fan inequality :
Before that we make the following substitution:
$x=frac{1}{a}$
$y=frac{1}{b}$
$z=frac{1}{c}$
We get the following inequality :
$$sum_{cyc}frac{(1-a)^2}{a^2}geq sum_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$$
Now we have the Ky-fan inequality wich state that for $a,b,c$ belongs to $[frac{1}{4};frac{1}{2}]$:
$$frac{(1-a)^2}{a^2}geqfrac{b^2c^2}{(1-b)^2(1-c)^2}Big(frac{3-a-b-c}{a+b+c}Big)^6
$$
So we get this refinement of the starting inequality if we sum each term of the previous inequality:
$$sum_{cyc}frac{(1-a)^2}{a^2}geq sum_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}Big(frac{3-a-b-c}{a+b+c}Big)^6 geqsum_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$$
So we prove the inequality for $a,b,c$ belongs to $[frac{1}{4};frac{1}{2}]$
To prove the remainder of the theorem you just have to play with a variable $u$ and remark that we always have for all positive $x,y,z$:
$$sum_{cyc}frac{(1-x)^2}{x^2}geq usum_{cyc}frac{(1-a)^2}{a^2}geq usum_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}Big(frac{3-a-b-c}{a+b+c}Big)^6\ geq usum_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}geq sum_{cyc}frac{(1-y^2)(1-z^2)}{(yz+x)^2}$$
We make the following substitution on the first inequality :
$x=cos^2 s$;
$y=cos^2 v$;
$z=cos^2 w$
And
$a=cos^2 p$;
$b=cos^2 r$;
$c=cos^2 q$
With $p,q,r$ belonging to $[arccos (frac{1}{sqrt{2}});arccos(frac{1}{2})]$ :
We get this :
$$sum_{cyc}tan^4 sgeq usum_{cyc}tan^4 p$$
we put :
$$u=frac{sum_{cyc}tan^2 stan^2 v}{sum_{cyc}tan^4 p}$$
Edit for Martin R :
We have build $u$ exactly to obtain the first inequality :
So we have :
$$usum_{cyc}tan^4 p=sum_{cyc}tan^2 stan^2 v$$
And
$$sum_{cyc}tan^4 sgeqsum_{cyc}tan^2 stan^2 v$$
And to conclude it's just Am-Gm or $a^2+b^2geq 2ab$
Now we need to prove that $u$ satisfy the last inequality :
With the previous substitution and combine with the well-know formula $(cos a^2+sin a^2)^2=1$ we find:
$$frac{sum_{cyc}tan^2 stan v^2}{sum_{cyc}tan^4 p}sum_{cyc}frac{(tan^4 p+2tan^2 p)((tan^4 r+2tan^2 r)}{Big(1+frac{cos^2 q}{cos^2 rcos^2 p}Big)^2}geqsum_{cyc}frac{(tan^4 s+2tan^2 s)((tan^4 v+2tan^2 v)}{Big(1+frac{cos^2 w}{cos^2 vcos^2 s}Big)^2}$$
edited Jul 6 '17 at 16:50
answered Jun 30 '17 at 19:17
user448747
1
$begingroup$
Why $sumlimits_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}left(frac{3-a-b-c}{a+b+c}right)^6 geqsumlimits_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$ is true?
$endgroup$
– Michael Rozenberg
Jul 3 '17 at 8:15
$begingroup$
See here.It works because there is no order on $a,b,c$.
$endgroup$
– user448747
Jul 3 '17 at 12:55
$begingroup$
Explain please your play with a variable $u$ for the rest cases.
$endgroup$
– Michael Rozenberg
Jul 3 '17 at 13:16
$begingroup$
If you use logarithmic majorization you will find a particular value for $u$ ;$u=frac{(1-x)^2a^2}{x^2(1-a)^2}$
$endgroup$
– user448747
Jul 3 '17 at 15:06
1
$begingroup$
Can someone, which upper voted this post, explain me why this solution is true? I think it's nothing.
$endgroup$
– Michael Rozenberg
Jul 6 '17 at 5:30
|
show 5 more comments
1
$begingroup$
Why $sumlimits_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}left(frac{3-a-b-c}{a+b+c}right)^6 geqsumlimits_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$ is true?
$endgroup$
– Michael Rozenberg
Jul 3 '17 at 8:15
$begingroup$
See here.It works because there is no order on $a,b,c$.
$endgroup$
– user448747
Jul 3 '17 at 12:55
$begingroup$
Explain please your play with a variable $u$ for the rest cases.
$endgroup$
– Michael Rozenberg
Jul 3 '17 at 13:16
$begingroup$
If you use logarithmic majorization you will find a particular value for $u$ ;$u=frac{(1-x)^2a^2}{x^2(1-a)^2}$
$endgroup$
– user448747
Jul 3 '17 at 15:06
1
$begingroup$
Can someone, which upper voted this post, explain me why this solution is true? I think it's nothing.
$endgroup$
– Michael Rozenberg
Jul 6 '17 at 5:30
1
1
$begingroup$
Why $sumlimits_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}left(frac{3-a-b-c}{a+b+c}right)^6 geqsumlimits_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$ is true?
$endgroup$
– Michael Rozenberg
Jul 3 '17 at 8:15
$begingroup$
Why $sumlimits_{cyc}frac{b^2c^2}{(1-b)^2(1-c)^2}left(frac{3-a-b-c}{a+b+c}right)^6 geqsumlimits_{cyc} frac{(1-a^2)(1-b^2)}{(ab+c)^2}$ is true?
$endgroup$
– Michael Rozenberg
Jul 3 '17 at 8:15
$begingroup$
See here.It works because there is no order on $a,b,c$.
$endgroup$
– user448747
Jul 3 '17 at 12:55
$begingroup$
See here.It works because there is no order on $a,b,c$.
$endgroup$
– user448747
Jul 3 '17 at 12:55
$begingroup$
Explain please your play with a variable $u$ for the rest cases.
$endgroup$
– Michael Rozenberg
Jul 3 '17 at 13:16
$begingroup$
Explain please your play with a variable $u$ for the rest cases.
$endgroup$
– Michael Rozenberg
Jul 3 '17 at 13:16
$begingroup$
If you use logarithmic majorization you will find a particular value for $u$ ;$u=frac{(1-x)^2a^2}{x^2(1-a)^2}$
$endgroup$
– user448747
Jul 3 '17 at 15:06
$begingroup$
If you use logarithmic majorization you will find a particular value for $u$ ;$u=frac{(1-x)^2a^2}{x^2(1-a)^2}$
$endgroup$
– user448747
Jul 3 '17 at 15:06
1
1
$begingroup$
Can someone, which upper voted this post, explain me why this solution is true? I think it's nothing.
$endgroup$
– Michael Rozenberg
Jul 6 '17 at 5:30
$begingroup$
Can someone, which upper voted this post, explain me why this solution is true? I think it's nothing.
$endgroup$
– Michael Rozenberg
Jul 6 '17 at 5:30
|
show 5 more comments
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$begingroup$
@Rozenberg Is inequality true for all nonnegatives? Or is there counter-exs ?
$endgroup$
– Merdanov
Jun 22 '17 at 12:00
$begingroup$
maybe let $y=x+p, z=x+p+q.p,qge 0$ is usefull?
$endgroup$
– communnites
Jun 29 '17 at 4:54
$begingroup$
I have a proof for $x+y+zgeqfrac{3}{2}.$
$endgroup$
– Michael Rozenberg
Dec 21 '18 at 13:01
$begingroup$
Hint became the answer.
$endgroup$
– Yuri Negometyanov
Dec 27 '18 at 14:49