show $sum_{cyc}(1-x)^2ge sum_{cyc}frac{z^2(1-x^2)(1-y^2)}{(xy+z)^2}.$












7












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    @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


















7












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    @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
















7












7








7


5



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 5 '17 at 11:11









Arjang

5,65462364




5,65462364










asked Jun 21 '17 at 0:56









communnitescommunnites

1,2431535




1,2431535












  • $begingroup$
    @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










  • $begingroup$
    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












1 Answer
1






active

oldest

votes


















2





+25







$begingroup$

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:



$$frac{sum_{cyc}tan^2 stan v^2}{sum_{cyc}tan^4 p}sum_{cyc}frac{(tan^4 p+2tan^2 p)((tan^4 r+2tan^2 r)}{Big(1+frac{cos^2 q}{cos^2 rcos^2 p}Big)^2}geqsum_{cyc}frac{(tan^4 s+2tan^2 s)((tan^4 v+2tan^2 v)}{Big(1+frac{cos^2 w}{cos^2 vcos^2 s}Big)^2}$$






share|cite|improve this answer











$endgroup$









  • 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












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2





+25







$begingroup$

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:



$$frac{sum_{cyc}tan^2 stan v^2}{sum_{cyc}tan^4 p}sum_{cyc}frac{(tan^4 p+2tan^2 p)((tan^4 r+2tan^2 r)}{Big(1+frac{cos^2 q}{cos^2 rcos^2 p}Big)^2}geqsum_{cyc}frac{(tan^4 s+2tan^2 s)((tan^4 v+2tan^2 v)}{Big(1+frac{cos^2 w}{cos^2 vcos^2 s}Big)^2}$$






share|cite|improve this answer











$endgroup$









  • 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
















2





+25







$begingroup$

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:



$$frac{sum_{cyc}tan^2 stan v^2}{sum_{cyc}tan^4 p}sum_{cyc}frac{(tan^4 p+2tan^2 p)((tan^4 r+2tan^2 r)}{Big(1+frac{cos^2 q}{cos^2 rcos^2 p}Big)^2}geqsum_{cyc}frac{(tan^4 s+2tan^2 s)((tan^4 v+2tan^2 v)}{Big(1+frac{cos^2 w}{cos^2 vcos^2 s}Big)^2}$$






share|cite|improve this answer











$endgroup$









  • 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














2





+25







2





+25



2




+25



$begingroup$

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:



$$frac{sum_{cyc}tan^2 stan v^2}{sum_{cyc}tan^4 p}sum_{cyc}frac{(tan^4 p+2tan^2 p)((tan^4 r+2tan^2 r)}{Big(1+frac{cos^2 q}{cos^2 rcos^2 p}Big)^2}geqsum_{cyc}frac{(tan^4 s+2tan^2 s)((tan^4 v+2tan^2 v)}{Big(1+frac{cos^2 w}{cos^2 vcos^2 s}Big)^2}$$






share|cite|improve this answer











$endgroup$



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:



$$frac{sum_{cyc}tan^2 stan v^2}{sum_{cyc}tan^4 p}sum_{cyc}frac{(tan^4 p+2tan^2 p)((tan^4 r+2tan^2 r)}{Big(1+frac{cos^2 q}{cos^2 rcos^2 p}Big)^2}geqsum_{cyc}frac{(tan^4 s+2tan^2 s)((tan^4 v+2tan^2 v)}{Big(1+frac{cos^2 w}{cos^2 vcos^2 s}Big)^2}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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














  • 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


















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