Inequality in 5 variables with conditions












2












$begingroup$


Let 5 positive real variables $(a,b,c,d,e)$. These variables obey the conditions $a-b+c-d+e > 0$ and all 5 cyclic shifts thereof. This question is tighter than this one.



Prove:
$$
sum_{cyc} a^2 b d (c+e)ge sum_{cyc} a b c e (a+d)
$$
where $sum_{cyc}$ means all 5 cyclic shifts $(a,b,c,d,e) to (b,c,d,e,a) to$ etc. Equality occurs if all 5 variables are equal, and it appears that equality occurs at no other points. I couldn't find counterexamples through simulations.






algebra-precalculus inequality







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$endgroup$








  • 1




    $begingroup$
    It would be interesting to know where that problem comes from.
    $endgroup$
    – Martin R
    Dec 12 '18 at 22:14
















2












$begingroup$


Let 5 positive real variables $(a,b,c,d,e)$. These variables obey the conditions $a-b+c-d+e > 0$ and all 5 cyclic shifts thereof. This question is tighter than this one.



Prove:
$$
sum_{cyc} a^2 b d (c+e)ge sum_{cyc} a b c e (a+d)
$$
where $sum_{cyc}$ means all 5 cyclic shifts $(a,b,c,d,e) to (b,c,d,e,a) to$ etc. Equality occurs if all 5 variables are equal, and it appears that equality occurs at no other points. I couldn't find counterexamples through simulations.






algebra-precalculus inequality







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It would be interesting to know where that problem comes from.
    $endgroup$
    – Martin R
    Dec 12 '18 at 22:14














2












2








2





$begingroup$


Let 5 positive real variables $(a,b,c,d,e)$. These variables obey the conditions $a-b+c-d+e > 0$ and all 5 cyclic shifts thereof. This question is tighter than this one.



Prove:
$$
sum_{cyc} a^2 b d (c+e)ge sum_{cyc} a b c e (a+d)
$$
where $sum_{cyc}$ means all 5 cyclic shifts $(a,b,c,d,e) to (b,c,d,e,a) to$ etc. Equality occurs if all 5 variables are equal, and it appears that equality occurs at no other points. I couldn't find counterexamples through simulations.






algebra-precalculus inequality







share|cite|improve this question









$endgroup$




Let 5 positive real variables $(a,b,c,d,e)$. These variables obey the conditions $a-b+c-d+e > 0$ and all 5 cyclic shifts thereof. This question is tighter than this one.



Prove:
$$
sum_{cyc} a^2 b d (c+e)ge sum_{cyc} a b c e (a+d)
$$
where $sum_{cyc}$ means all 5 cyclic shifts $(a,b,c,d,e) to (b,c,d,e,a) to$ etc. Equality occurs if all 5 variables are equal, and it appears that equality occurs at no other points. I couldn't find counterexamples through simulations.






algebra-precalculus inequality




algebra-precalculus inequality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 12 '18 at 22:00









AndreasAndreas

8,4161137




8,4161137








  • 1




    $begingroup$
    It would be interesting to know where that problem comes from.
    $endgroup$
    – Martin R
    Dec 12 '18 at 22:14














  • 1




    $begingroup$
    It would be interesting to know where that problem comes from.
    $endgroup$
    – Martin R
    Dec 12 '18 at 22:14








1




1




$begingroup$
It would be interesting to know where that problem comes from.
$endgroup$
– Martin R
Dec 12 '18 at 22:14




$begingroup$
It would be interesting to know where that problem comes from.
$endgroup$
– Martin R
Dec 12 '18 at 22:14










1 Answer
1






active

oldest

votes


















0












$begingroup$

The hint.



Let $a=x+y$, $b=y+z$, $c=z+t$, $d=t+w$ and $e=w+x$.



Thus, the condition gives that $x$, $y$, $z$, $t$ and $w$ are positives and we need to prove that
$$sum_{cyc}(x+y)^2(y+z)(t^2+w^2+zt+tw+wx-zx)geq5prod_{cyc}(x+y),$$ which we can prove by BW:



Let $x=min{x,y,z,t,w}$, $y=x+p$, $z=x+q$, $t=x+r$ and $w=x+s.$



Practically it's better to use the last substitution for the starting inequality:
$$sum_{cyc}a^2b(cd+de-ce)geq5abcde.$$
Id est, let $a=x+u,$ $b=x+u+v$, $c=x+v+w$, $d=x+w+p$ and $e=x+p$,



where $x>0$ and $u$, $v$, $w$ and $p$ are non-negatives.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Here is the result for $sum_{cyc}a^2b(cd+de-ce)- 5abcde geq 0 $ with BW and it doesn't look to me like it it easy to prove that it is nonnegative
    $endgroup$
    – Andreas
    Dec 13 '18 at 19:45










  • $begingroup$
    $p^3*u*v + p^3*u*w + p^3*u + p^3*v^2 + p^3*v*w + 2*p^3*v + p^3*w + p^3 + p^2*u^3 - 2*p^2*u^2*v - 2*p^2*u^2*w - 2*p^2*u*v^2 - 3*p^2*u*v + p^2*u*w^2 + p^2*v^2*w + p^2*v*w^2 + 3*p^2*v*w + p^2*v + p^2*w^2 + 2*p^2*w + p^2 + p*u^3*w + 2*p*u^3 + p*u^2*v^2 - p*u^2*v*w - 2*p*u^2*v - 2*p*u^2*w^2 - 3*p*u^2*w + p*u^2 + p*u*v^3 - p*u*v^2*w - p*u*v*w^2 - 3*p*u*v*w - 4*p*u*v - 2*p*u*w^2 - 4*p*u*w - p*u + p*v^3 - p*v + u^3*w + u^3 + u^2*v^2 + u^2*v*w^2 + u^2*w^3 + u^2 + u*v^3 + u*v^2*w^2 + 2*u*v^2 + u*v*w^3 + 3*u*v*w^2 + 2*u*w^3 + u*w^2 - u*w + v^3 + v^2*w^2 + v^2 + v*w^3 + 2*v*w^2 + w^3 + w^2$
    $endgroup$
    – Andreas
    Dec 13 '18 at 19:45










  • $begingroup$
    @Andreas It should be homogeneous inequality of degree five, which says that there is a mistake in your computations (I have no some software and If you just wish I can calculate by hand. But it's better- no :)) You must get $A(u,v,w,p)x^3+B(u,v,w,p)x^2+C(u,v,w,p)x+D(u,v,w,p)geq0$, where $A$ is a homogeneous second degree polynomial, $B$ is a homogeneous third degree polynomial, $C$ is a homogeneous fourth degree polynomial and $D$ is a homogeneous fifth degree polynomial. It seems that we'll get $Ageq0,$ $Bgeq0$, $Cgeq0$ and $Dgeq0.$ Maybe did you put $x=1$? If so, we are done!
    $endgroup$
    – Michael Rozenberg
    Dec 13 '18 at 20:59












  • $begingroup$
    You're right, I put x= 1. Here is the result with free x: $x^3(p^2 - pu - pv + u^2 - uw + v^2 + w^2) + x^2(p^3 + p^2v + 2p^2w + pu^2 - 4puv - 4puw + u^3 + 2uv^2 + uw^2 + v^3 + 2vw^2 + w^3) + x(p^3u + 2p^3v + p^3w - 3p^2uv + 3p^2vw + p^2w^2 + 2pu^3 - 2pu^2v - 3pu^2w - 3puvw - 2puw^2 + pv^3 + u^3w + u^2v^2 + uv^3 + 3uvw^2 + 2uw^3 + v^2w^2 + vw^3) + p^3uv + p^3uw + p^3v^2 + p^3vw + p^2u^3 - 2p^2u^2v - 2p^2u^2w - 2p^2uv^2 + p^2uw^2 + p^2v^2w + p^2vw^2 + pu^3w + pu^2v^2 - pu^2vw - 2pu^2w^2 + puv^3 - puv^2w - puvw^2 + u^2vw^2 + u^2w^3 + uv^2w^2 + uvw^3 $
    $endgroup$
    – Andreas
    Dec 14 '18 at 14:52












  • $begingroup$
    Using $pu le frac12 (p^2 + u^2)$ etc., the first polynomial is $A ge frac12 (v^2 + w^2) ge 0$. The next ones $B, C, D$ get more complicated.
    $endgroup$
    – Andreas
    Dec 14 '18 at 15:23












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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

The hint.



Let $a=x+y$, $b=y+z$, $c=z+t$, $d=t+w$ and $e=w+x$.



Thus, the condition gives that $x$, $y$, $z$, $t$ and $w$ are positives and we need to prove that
$$sum_{cyc}(x+y)^2(y+z)(t^2+w^2+zt+tw+wx-zx)geq5prod_{cyc}(x+y),$$ which we can prove by BW:



Let $x=min{x,y,z,t,w}$, $y=x+p$, $z=x+q$, $t=x+r$ and $w=x+s.$



Practically it's better to use the last substitution for the starting inequality:
$$sum_{cyc}a^2b(cd+de-ce)geq5abcde.$$
Id est, let $a=x+u,$ $b=x+u+v$, $c=x+v+w$, $d=x+w+p$ and $e=x+p$,



where $x>0$ and $u$, $v$, $w$ and $p$ are non-negatives.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Here is the result for $sum_{cyc}a^2b(cd+de-ce)- 5abcde geq 0 $ with BW and it doesn't look to me like it it easy to prove that it is nonnegative
    $endgroup$
    – Andreas
    Dec 13 '18 at 19:45










  • $begingroup$
    $p^3*u*v + p^3*u*w + p^3*u + p^3*v^2 + p^3*v*w + 2*p^3*v + p^3*w + p^3 + p^2*u^3 - 2*p^2*u^2*v - 2*p^2*u^2*w - 2*p^2*u*v^2 - 3*p^2*u*v + p^2*u*w^2 + p^2*v^2*w + p^2*v*w^2 + 3*p^2*v*w + p^2*v + p^2*w^2 + 2*p^2*w + p^2 + p*u^3*w + 2*p*u^3 + p*u^2*v^2 - p*u^2*v*w - 2*p*u^2*v - 2*p*u^2*w^2 - 3*p*u^2*w + p*u^2 + p*u*v^3 - p*u*v^2*w - p*u*v*w^2 - 3*p*u*v*w - 4*p*u*v - 2*p*u*w^2 - 4*p*u*w - p*u + p*v^3 - p*v + u^3*w + u^3 + u^2*v^2 + u^2*v*w^2 + u^2*w^3 + u^2 + u*v^3 + u*v^2*w^2 + 2*u*v^2 + u*v*w^3 + 3*u*v*w^2 + 2*u*w^3 + u*w^2 - u*w + v^3 + v^2*w^2 + v^2 + v*w^3 + 2*v*w^2 + w^3 + w^2$
    $endgroup$
    – Andreas
    Dec 13 '18 at 19:45










  • $begingroup$
    @Andreas It should be homogeneous inequality of degree five, which says that there is a mistake in your computations (I have no some software and If you just wish I can calculate by hand. But it's better- no :)) You must get $A(u,v,w,p)x^3+B(u,v,w,p)x^2+C(u,v,w,p)x+D(u,v,w,p)geq0$, where $A$ is a homogeneous second degree polynomial, $B$ is a homogeneous third degree polynomial, $C$ is a homogeneous fourth degree polynomial and $D$ is a homogeneous fifth degree polynomial. It seems that we'll get $Ageq0,$ $Bgeq0$, $Cgeq0$ and $Dgeq0.$ Maybe did you put $x=1$? If so, we are done!
    $endgroup$
    – Michael Rozenberg
    Dec 13 '18 at 20:59












  • $begingroup$
    You're right, I put x= 1. Here is the result with free x: $x^3(p^2 - pu - pv + u^2 - uw + v^2 + w^2) + x^2(p^3 + p^2v + 2p^2w + pu^2 - 4puv - 4puw + u^3 + 2uv^2 + uw^2 + v^3 + 2vw^2 + w^3) + x(p^3u + 2p^3v + p^3w - 3p^2uv + 3p^2vw + p^2w^2 + 2pu^3 - 2pu^2v - 3pu^2w - 3puvw - 2puw^2 + pv^3 + u^3w + u^2v^2 + uv^3 + 3uvw^2 + 2uw^3 + v^2w^2 + vw^3) + p^3uv + p^3uw + p^3v^2 + p^3vw + p^2u^3 - 2p^2u^2v - 2p^2u^2w - 2p^2uv^2 + p^2uw^2 + p^2v^2w + p^2vw^2 + pu^3w + pu^2v^2 - pu^2vw - 2pu^2w^2 + puv^3 - puv^2w - puvw^2 + u^2vw^2 + u^2w^3 + uv^2w^2 + uvw^3 $
    $endgroup$
    – Andreas
    Dec 14 '18 at 14:52












  • $begingroup$
    Using $pu le frac12 (p^2 + u^2)$ etc., the first polynomial is $A ge frac12 (v^2 + w^2) ge 0$. The next ones $B, C, D$ get more complicated.
    $endgroup$
    – Andreas
    Dec 14 '18 at 15:23
















0












$begingroup$

The hint.



Let $a=x+y$, $b=y+z$, $c=z+t$, $d=t+w$ and $e=w+x$.



Thus, the condition gives that $x$, $y$, $z$, $t$ and $w$ are positives and we need to prove that
$$sum_{cyc}(x+y)^2(y+z)(t^2+w^2+zt+tw+wx-zx)geq5prod_{cyc}(x+y),$$ which we can prove by BW:



Let $x=min{x,y,z,t,w}$, $y=x+p$, $z=x+q$, $t=x+r$ and $w=x+s.$



Practically it's better to use the last substitution for the starting inequality:
$$sum_{cyc}a^2b(cd+de-ce)geq5abcde.$$
Id est, let $a=x+u,$ $b=x+u+v$, $c=x+v+w$, $d=x+w+p$ and $e=x+p$,



where $x>0$ and $u$, $v$, $w$ and $p$ are non-negatives.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Here is the result for $sum_{cyc}a^2b(cd+de-ce)- 5abcde geq 0 $ with BW and it doesn't look to me like it it easy to prove that it is nonnegative
    $endgroup$
    – Andreas
    Dec 13 '18 at 19:45










  • $begingroup$
    $p^3*u*v + p^3*u*w + p^3*u + p^3*v^2 + p^3*v*w + 2*p^3*v + p^3*w + p^3 + p^2*u^3 - 2*p^2*u^2*v - 2*p^2*u^2*w - 2*p^2*u*v^2 - 3*p^2*u*v + p^2*u*w^2 + p^2*v^2*w + p^2*v*w^2 + 3*p^2*v*w + p^2*v + p^2*w^2 + 2*p^2*w + p^2 + p*u^3*w + 2*p*u^3 + p*u^2*v^2 - p*u^2*v*w - 2*p*u^2*v - 2*p*u^2*w^2 - 3*p*u^2*w + p*u^2 + p*u*v^3 - p*u*v^2*w - p*u*v*w^2 - 3*p*u*v*w - 4*p*u*v - 2*p*u*w^2 - 4*p*u*w - p*u + p*v^3 - p*v + u^3*w + u^3 + u^2*v^2 + u^2*v*w^2 + u^2*w^3 + u^2 + u*v^3 + u*v^2*w^2 + 2*u*v^2 + u*v*w^3 + 3*u*v*w^2 + 2*u*w^3 + u*w^2 - u*w + v^3 + v^2*w^2 + v^2 + v*w^3 + 2*v*w^2 + w^3 + w^2$
    $endgroup$
    – Andreas
    Dec 13 '18 at 19:45










  • $begingroup$
    @Andreas It should be homogeneous inequality of degree five, which says that there is a mistake in your computations (I have no some software and If you just wish I can calculate by hand. But it's better- no :)) You must get $A(u,v,w,p)x^3+B(u,v,w,p)x^2+C(u,v,w,p)x+D(u,v,w,p)geq0$, where $A$ is a homogeneous second degree polynomial, $B$ is a homogeneous third degree polynomial, $C$ is a homogeneous fourth degree polynomial and $D$ is a homogeneous fifth degree polynomial. It seems that we'll get $Ageq0,$ $Bgeq0$, $Cgeq0$ and $Dgeq0.$ Maybe did you put $x=1$? If so, we are done!
    $endgroup$
    – Michael Rozenberg
    Dec 13 '18 at 20:59












  • $begingroup$
    You're right, I put x= 1. Here is the result with free x: $x^3(p^2 - pu - pv + u^2 - uw + v^2 + w^2) + x^2(p^3 + p^2v + 2p^2w + pu^2 - 4puv - 4puw + u^3 + 2uv^2 + uw^2 + v^3 + 2vw^2 + w^3) + x(p^3u + 2p^3v + p^3w - 3p^2uv + 3p^2vw + p^2w^2 + 2pu^3 - 2pu^2v - 3pu^2w - 3puvw - 2puw^2 + pv^3 + u^3w + u^2v^2 + uv^3 + 3uvw^2 + 2uw^3 + v^2w^2 + vw^3) + p^3uv + p^3uw + p^3v^2 + p^3vw + p^2u^3 - 2p^2u^2v - 2p^2u^2w - 2p^2uv^2 + p^2uw^2 + p^2v^2w + p^2vw^2 + pu^3w + pu^2v^2 - pu^2vw - 2pu^2w^2 + puv^3 - puv^2w - puvw^2 + u^2vw^2 + u^2w^3 + uv^2w^2 + uvw^3 $
    $endgroup$
    – Andreas
    Dec 14 '18 at 14:52












  • $begingroup$
    Using $pu le frac12 (p^2 + u^2)$ etc., the first polynomial is $A ge frac12 (v^2 + w^2) ge 0$. The next ones $B, C, D$ get more complicated.
    $endgroup$
    – Andreas
    Dec 14 '18 at 15:23














0












0








0





$begingroup$

The hint.



Let $a=x+y$, $b=y+z$, $c=z+t$, $d=t+w$ and $e=w+x$.



Thus, the condition gives that $x$, $y$, $z$, $t$ and $w$ are positives and we need to prove that
$$sum_{cyc}(x+y)^2(y+z)(t^2+w^2+zt+tw+wx-zx)geq5prod_{cyc}(x+y),$$ which we can prove by BW:



Let $x=min{x,y,z,t,w}$, $y=x+p$, $z=x+q$, $t=x+r$ and $w=x+s.$



Practically it's better to use the last substitution for the starting inequality:
$$sum_{cyc}a^2b(cd+de-ce)geq5abcde.$$
Id est, let $a=x+u,$ $b=x+u+v$, $c=x+v+w$, $d=x+w+p$ and $e=x+p$,



where $x>0$ and $u$, $v$, $w$ and $p$ are non-negatives.






share|cite|improve this answer











$endgroup$



The hint.



Let $a=x+y$, $b=y+z$, $c=z+t$, $d=t+w$ and $e=w+x$.



Thus, the condition gives that $x$, $y$, $z$, $t$ and $w$ are positives and we need to prove that
$$sum_{cyc}(x+y)^2(y+z)(t^2+w^2+zt+tw+wx-zx)geq5prod_{cyc}(x+y),$$ which we can prove by BW:



Let $x=min{x,y,z,t,w}$, $y=x+p$, $z=x+q$, $t=x+r$ and $w=x+s.$



Practically it's better to use the last substitution for the starting inequality:
$$sum_{cyc}a^2b(cd+de-ce)geq5abcde.$$
Id est, let $a=x+u,$ $b=x+u+v$, $c=x+v+w$, $d=x+w+p$ and $e=x+p$,



where $x>0$ and $u$, $v$, $w$ and $p$ are non-negatives.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 12 '18 at 23:45

























answered Dec 12 '18 at 22:29









Michael RozenbergMichael Rozenberg

109k1896201




109k1896201












  • $begingroup$
    Here is the result for $sum_{cyc}a^2b(cd+de-ce)- 5abcde geq 0 $ with BW and it doesn't look to me like it it easy to prove that it is nonnegative
    $endgroup$
    – Andreas
    Dec 13 '18 at 19:45










  • $begingroup$
    $p^3*u*v + p^3*u*w + p^3*u + p^3*v^2 + p^3*v*w + 2*p^3*v + p^3*w + p^3 + p^2*u^3 - 2*p^2*u^2*v - 2*p^2*u^2*w - 2*p^2*u*v^2 - 3*p^2*u*v + p^2*u*w^2 + p^2*v^2*w + p^2*v*w^2 + 3*p^2*v*w + p^2*v + p^2*w^2 + 2*p^2*w + p^2 + p*u^3*w + 2*p*u^3 + p*u^2*v^2 - p*u^2*v*w - 2*p*u^2*v - 2*p*u^2*w^2 - 3*p*u^2*w + p*u^2 + p*u*v^3 - p*u*v^2*w - p*u*v*w^2 - 3*p*u*v*w - 4*p*u*v - 2*p*u*w^2 - 4*p*u*w - p*u + p*v^3 - p*v + u^3*w + u^3 + u^2*v^2 + u^2*v*w^2 + u^2*w^3 + u^2 + u*v^3 + u*v^2*w^2 + 2*u*v^2 + u*v*w^3 + 3*u*v*w^2 + 2*u*w^3 + u*w^2 - u*w + v^3 + v^2*w^2 + v^2 + v*w^3 + 2*v*w^2 + w^3 + w^2$
    $endgroup$
    – Andreas
    Dec 13 '18 at 19:45










  • $begingroup$
    @Andreas It should be homogeneous inequality of degree five, which says that there is a mistake in your computations (I have no some software and If you just wish I can calculate by hand. But it's better- no :)) You must get $A(u,v,w,p)x^3+B(u,v,w,p)x^2+C(u,v,w,p)x+D(u,v,w,p)geq0$, where $A$ is a homogeneous second degree polynomial, $B$ is a homogeneous third degree polynomial, $C$ is a homogeneous fourth degree polynomial and $D$ is a homogeneous fifth degree polynomial. It seems that we'll get $Ageq0,$ $Bgeq0$, $Cgeq0$ and $Dgeq0.$ Maybe did you put $x=1$? If so, we are done!
    $endgroup$
    – Michael Rozenberg
    Dec 13 '18 at 20:59












  • $begingroup$
    You're right, I put x= 1. Here is the result with free x: $x^3(p^2 - pu - pv + u^2 - uw + v^2 + w^2) + x^2(p^3 + p^2v + 2p^2w + pu^2 - 4puv - 4puw + u^3 + 2uv^2 + uw^2 + v^3 + 2vw^2 + w^3) + x(p^3u + 2p^3v + p^3w - 3p^2uv + 3p^2vw + p^2w^2 + 2pu^3 - 2pu^2v - 3pu^2w - 3puvw - 2puw^2 + pv^3 + u^3w + u^2v^2 + uv^3 + 3uvw^2 + 2uw^3 + v^2w^2 + vw^3) + p^3uv + p^3uw + p^3v^2 + p^3vw + p^2u^3 - 2p^2u^2v - 2p^2u^2w - 2p^2uv^2 + p^2uw^2 + p^2v^2w + p^2vw^2 + pu^3w + pu^2v^2 - pu^2vw - 2pu^2w^2 + puv^3 - puv^2w - puvw^2 + u^2vw^2 + u^2w^3 + uv^2w^2 + uvw^3 $
    $endgroup$
    – Andreas
    Dec 14 '18 at 14:52












  • $begingroup$
    Using $pu le frac12 (p^2 + u^2)$ etc., the first polynomial is $A ge frac12 (v^2 + w^2) ge 0$. The next ones $B, C, D$ get more complicated.
    $endgroup$
    – Andreas
    Dec 14 '18 at 15:23


















  • $begingroup$
    Here is the result for $sum_{cyc}a^2b(cd+de-ce)- 5abcde geq 0 $ with BW and it doesn't look to me like it it easy to prove that it is nonnegative
    $endgroup$
    – Andreas
    Dec 13 '18 at 19:45










  • $begingroup$
    $p^3*u*v + p^3*u*w + p^3*u + p^3*v^2 + p^3*v*w + 2*p^3*v + p^3*w + p^3 + p^2*u^3 - 2*p^2*u^2*v - 2*p^2*u^2*w - 2*p^2*u*v^2 - 3*p^2*u*v + p^2*u*w^2 + p^2*v^2*w + p^2*v*w^2 + 3*p^2*v*w + p^2*v + p^2*w^2 + 2*p^2*w + p^2 + p*u^3*w + 2*p*u^3 + p*u^2*v^2 - p*u^2*v*w - 2*p*u^2*v - 2*p*u^2*w^2 - 3*p*u^2*w + p*u^2 + p*u*v^3 - p*u*v^2*w - p*u*v*w^2 - 3*p*u*v*w - 4*p*u*v - 2*p*u*w^2 - 4*p*u*w - p*u + p*v^3 - p*v + u^3*w + u^3 + u^2*v^2 + u^2*v*w^2 + u^2*w^3 + u^2 + u*v^3 + u*v^2*w^2 + 2*u*v^2 + u*v*w^3 + 3*u*v*w^2 + 2*u*w^3 + u*w^2 - u*w + v^3 + v^2*w^2 + v^2 + v*w^3 + 2*v*w^2 + w^3 + w^2$
    $endgroup$
    – Andreas
    Dec 13 '18 at 19:45










  • $begingroup$
    @Andreas It should be homogeneous inequality of degree five, which says that there is a mistake in your computations (I have no some software and If you just wish I can calculate by hand. But it's better- no :)) You must get $A(u,v,w,p)x^3+B(u,v,w,p)x^2+C(u,v,w,p)x+D(u,v,w,p)geq0$, where $A$ is a homogeneous second degree polynomial, $B$ is a homogeneous third degree polynomial, $C$ is a homogeneous fourth degree polynomial and $D$ is a homogeneous fifth degree polynomial. It seems that we'll get $Ageq0,$ $Bgeq0$, $Cgeq0$ and $Dgeq0.$ Maybe did you put $x=1$? If so, we are done!
    $endgroup$
    – Michael Rozenberg
    Dec 13 '18 at 20:59












  • $begingroup$
    You're right, I put x= 1. Here is the result with free x: $x^3(p^2 - pu - pv + u^2 - uw + v^2 + w^2) + x^2(p^3 + p^2v + 2p^2w + pu^2 - 4puv - 4puw + u^3 + 2uv^2 + uw^2 + v^3 + 2vw^2 + w^3) + x(p^3u + 2p^3v + p^3w - 3p^2uv + 3p^2vw + p^2w^2 + 2pu^3 - 2pu^2v - 3pu^2w - 3puvw - 2puw^2 + pv^3 + u^3w + u^2v^2 + uv^3 + 3uvw^2 + 2uw^3 + v^2w^2 + vw^3) + p^3uv + p^3uw + p^3v^2 + p^3vw + p^2u^3 - 2p^2u^2v - 2p^2u^2w - 2p^2uv^2 + p^2uw^2 + p^2v^2w + p^2vw^2 + pu^3w + pu^2v^2 - pu^2vw - 2pu^2w^2 + puv^3 - puv^2w - puvw^2 + u^2vw^2 + u^2w^3 + uv^2w^2 + uvw^3 $
    $endgroup$
    – Andreas
    Dec 14 '18 at 14:52












  • $begingroup$
    Using $pu le frac12 (p^2 + u^2)$ etc., the first polynomial is $A ge frac12 (v^2 + w^2) ge 0$. The next ones $B, C, D$ get more complicated.
    $endgroup$
    – Andreas
    Dec 14 '18 at 15:23
















$begingroup$
Here is the result for $sum_{cyc}a^2b(cd+de-ce)- 5abcde geq 0 $ with BW and it doesn't look to me like it it easy to prove that it is nonnegative
$endgroup$
– Andreas
Dec 13 '18 at 19:45




$begingroup$
Here is the result for $sum_{cyc}a^2b(cd+de-ce)- 5abcde geq 0 $ with BW and it doesn't look to me like it it easy to prove that it is nonnegative
$endgroup$
– Andreas
Dec 13 '18 at 19:45












$begingroup$
$p^3*u*v + p^3*u*w + p^3*u + p^3*v^2 + p^3*v*w + 2*p^3*v + p^3*w + p^3 + p^2*u^3 - 2*p^2*u^2*v - 2*p^2*u^2*w - 2*p^2*u*v^2 - 3*p^2*u*v + p^2*u*w^2 + p^2*v^2*w + p^2*v*w^2 + 3*p^2*v*w + p^2*v + p^2*w^2 + 2*p^2*w + p^2 + p*u^3*w + 2*p*u^3 + p*u^2*v^2 - p*u^2*v*w - 2*p*u^2*v - 2*p*u^2*w^2 - 3*p*u^2*w + p*u^2 + p*u*v^3 - p*u*v^2*w - p*u*v*w^2 - 3*p*u*v*w - 4*p*u*v - 2*p*u*w^2 - 4*p*u*w - p*u + p*v^3 - p*v + u^3*w + u^3 + u^2*v^2 + u^2*v*w^2 + u^2*w^3 + u^2 + u*v^3 + u*v^2*w^2 + 2*u*v^2 + u*v*w^3 + 3*u*v*w^2 + 2*u*w^3 + u*w^2 - u*w + v^3 + v^2*w^2 + v^2 + v*w^3 + 2*v*w^2 + w^3 + w^2$
$endgroup$
– Andreas
Dec 13 '18 at 19:45




$begingroup$
$p^3*u*v + p^3*u*w + p^3*u + p^3*v^2 + p^3*v*w + 2*p^3*v + p^3*w + p^3 + p^2*u^3 - 2*p^2*u^2*v - 2*p^2*u^2*w - 2*p^2*u*v^2 - 3*p^2*u*v + p^2*u*w^2 + p^2*v^2*w + p^2*v*w^2 + 3*p^2*v*w + p^2*v + p^2*w^2 + 2*p^2*w + p^2 + p*u^3*w + 2*p*u^3 + p*u^2*v^2 - p*u^2*v*w - 2*p*u^2*v - 2*p*u^2*w^2 - 3*p*u^2*w + p*u^2 + p*u*v^3 - p*u*v^2*w - p*u*v*w^2 - 3*p*u*v*w - 4*p*u*v - 2*p*u*w^2 - 4*p*u*w - p*u + p*v^3 - p*v + u^3*w + u^3 + u^2*v^2 + u^2*v*w^2 + u^2*w^3 + u^2 + u*v^3 + u*v^2*w^2 + 2*u*v^2 + u*v*w^3 + 3*u*v*w^2 + 2*u*w^3 + u*w^2 - u*w + v^3 + v^2*w^2 + v^2 + v*w^3 + 2*v*w^2 + w^3 + w^2$
$endgroup$
– Andreas
Dec 13 '18 at 19:45












$begingroup$
@Andreas It should be homogeneous inequality of degree five, which says that there is a mistake in your computations (I have no some software and If you just wish I can calculate by hand. But it's better- no :)) You must get $A(u,v,w,p)x^3+B(u,v,w,p)x^2+C(u,v,w,p)x+D(u,v,w,p)geq0$, where $A$ is a homogeneous second degree polynomial, $B$ is a homogeneous third degree polynomial, $C$ is a homogeneous fourth degree polynomial and $D$ is a homogeneous fifth degree polynomial. It seems that we'll get $Ageq0,$ $Bgeq0$, $Cgeq0$ and $Dgeq0.$ Maybe did you put $x=1$? If so, we are done!
$endgroup$
– Michael Rozenberg
Dec 13 '18 at 20:59






$begingroup$
@Andreas It should be homogeneous inequality of degree five, which says that there is a mistake in your computations (I have no some software and If you just wish I can calculate by hand. But it's better- no :)) You must get $A(u,v,w,p)x^3+B(u,v,w,p)x^2+C(u,v,w,p)x+D(u,v,w,p)geq0$, where $A$ is a homogeneous second degree polynomial, $B$ is a homogeneous third degree polynomial, $C$ is a homogeneous fourth degree polynomial and $D$ is a homogeneous fifth degree polynomial. It seems that we'll get $Ageq0,$ $Bgeq0$, $Cgeq0$ and $Dgeq0.$ Maybe did you put $x=1$? If so, we are done!
$endgroup$
– Michael Rozenberg
Dec 13 '18 at 20:59














$begingroup$
You're right, I put x= 1. Here is the result with free x: $x^3(p^2 - pu - pv + u^2 - uw + v^2 + w^2) + x^2(p^3 + p^2v + 2p^2w + pu^2 - 4puv - 4puw + u^3 + 2uv^2 + uw^2 + v^3 + 2vw^2 + w^3) + x(p^3u + 2p^3v + p^3w - 3p^2uv + 3p^2vw + p^2w^2 + 2pu^3 - 2pu^2v - 3pu^2w - 3puvw - 2puw^2 + pv^3 + u^3w + u^2v^2 + uv^3 + 3uvw^2 + 2uw^3 + v^2w^2 + vw^3) + p^3uv + p^3uw + p^3v^2 + p^3vw + p^2u^3 - 2p^2u^2v - 2p^2u^2w - 2p^2uv^2 + p^2uw^2 + p^2v^2w + p^2vw^2 + pu^3w + pu^2v^2 - pu^2vw - 2pu^2w^2 + puv^3 - puv^2w - puvw^2 + u^2vw^2 + u^2w^3 + uv^2w^2 + uvw^3 $
$endgroup$
– Andreas
Dec 14 '18 at 14:52






$begingroup$
You're right, I put x= 1. Here is the result with free x: $x^3(p^2 - pu - pv + u^2 - uw + v^2 + w^2) + x^2(p^3 + p^2v + 2p^2w + pu^2 - 4puv - 4puw + u^3 + 2uv^2 + uw^2 + v^3 + 2vw^2 + w^3) + x(p^3u + 2p^3v + p^3w - 3p^2uv + 3p^2vw + p^2w^2 + 2pu^3 - 2pu^2v - 3pu^2w - 3puvw - 2puw^2 + pv^3 + u^3w + u^2v^2 + uv^3 + 3uvw^2 + 2uw^3 + v^2w^2 + vw^3) + p^3uv + p^3uw + p^3v^2 + p^3vw + p^2u^3 - 2p^2u^2v - 2p^2u^2w - 2p^2uv^2 + p^2uw^2 + p^2v^2w + p^2vw^2 + pu^3w + pu^2v^2 - pu^2vw - 2pu^2w^2 + puv^3 - puv^2w - puvw^2 + u^2vw^2 + u^2w^3 + uv^2w^2 + uvw^3 $
$endgroup$
– Andreas
Dec 14 '18 at 14:52














$begingroup$
Using $pu le frac12 (p^2 + u^2)$ etc., the first polynomial is $A ge frac12 (v^2 + w^2) ge 0$. The next ones $B, C, D$ get more complicated.
$endgroup$
– Andreas
Dec 14 '18 at 15:23




$begingroup$
Using $pu le frac12 (p^2 + u^2)$ etc., the first polynomial is $A ge frac12 (v^2 + w^2) ge 0$. The next ones $B, C, D$ get more complicated.
$endgroup$
– Andreas
Dec 14 '18 at 15:23


















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