Is this functional non-negative for $x_igeq 0$?
$begingroup$
Consider
$$
Phi(x)=sum_{i=1}^n a_i^2 (1+x_i)^2sum_{j=1}^n a_j^2 (1+x_j)(x_j-x_i)(x_1-x_i(2+x_j)),quad a_iinmathbb R.
$$
My conjecture is that this functional is non-negative for all $n$ and $x_igeq 0$. Is it true?
My approach is to put $x=t y$, where $yinmathbb R^n$ is a unit non-negative vector and then rewrite $Phi$ as a 6th-degree polynomial of $t$. Then I try to show that all its coefficients are positive (which is of course not a necessary condition). So far I think I have shown that the coefficients of $t^6$ and $t^5$ are always positive (using induction on $n$). The other coefficients are a bit clumsier and I wonder if there's a simpler way to prove the conjecture.
inequality
$endgroup$
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$begingroup$
Consider
$$
Phi(x)=sum_{i=1}^n a_i^2 (1+x_i)^2sum_{j=1}^n a_j^2 (1+x_j)(x_j-x_i)(x_1-x_i(2+x_j)),quad a_iinmathbb R.
$$
My conjecture is that this functional is non-negative for all $n$ and $x_igeq 0$. Is it true?
My approach is to put $x=t y$, where $yinmathbb R^n$ is a unit non-negative vector and then rewrite $Phi$ as a 6th-degree polynomial of $t$. Then I try to show that all its coefficients are positive (which is of course not a necessary condition). So far I think I have shown that the coefficients of $t^6$ and $t^5$ are always positive (using induction on $n$). The other coefficients are a bit clumsier and I wonder if there's a simpler way to prove the conjecture.
inequality
$endgroup$
add a comment |
$begingroup$
Consider
$$
Phi(x)=sum_{i=1}^n a_i^2 (1+x_i)^2sum_{j=1}^n a_j^2 (1+x_j)(x_j-x_i)(x_1-x_i(2+x_j)),quad a_iinmathbb R.
$$
My conjecture is that this functional is non-negative for all $n$ and $x_igeq 0$. Is it true?
My approach is to put $x=t y$, where $yinmathbb R^n$ is a unit non-negative vector and then rewrite $Phi$ as a 6th-degree polynomial of $t$. Then I try to show that all its coefficients are positive (which is of course not a necessary condition). So far I think I have shown that the coefficients of $t^6$ and $t^5$ are always positive (using induction on $n$). The other coefficients are a bit clumsier and I wonder if there's a simpler way to prove the conjecture.
inequality
$endgroup$
Consider
$$
Phi(x)=sum_{i=1}^n a_i^2 (1+x_i)^2sum_{j=1}^n a_j^2 (1+x_j)(x_j-x_i)(x_1-x_i(2+x_j)),quad a_iinmathbb R.
$$
My conjecture is that this functional is non-negative for all $n$ and $x_igeq 0$. Is it true?
My approach is to put $x=t y$, where $yinmathbb R^n$ is a unit non-negative vector and then rewrite $Phi$ as a 6th-degree polynomial of $t$. Then I try to show that all its coefficients are positive (which is of course not a necessary condition). So far I think I have shown that the coefficients of $t^6$ and $t^5$ are always positive (using induction on $n$). The other coefficients are a bit clumsier and I wonder if there's a simpler way to prove the conjecture.
inequality
inequality
edited Jan 15 at 22:29
Nasa Momdele
asked Jan 15 at 22:10
Nasa MomdeleNasa Momdele
254
254
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1 Answer
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$begingroup$
Random counterexample: $Phi(x)=-288$ when $n=3,,(a_1,a_2,a_3)=(0,4,3)$ and $x=(5,0,1)$.
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$begingroup$
Random counterexample: $Phi(x)=-288$ when $n=3,,(a_1,a_2,a_3)=(0,4,3)$ and $x=(5,0,1)$.
$endgroup$
add a comment |
$begingroup$
Random counterexample: $Phi(x)=-288$ when $n=3,,(a_1,a_2,a_3)=(0,4,3)$ and $x=(5,0,1)$.
$endgroup$
add a comment |
$begingroup$
Random counterexample: $Phi(x)=-288$ when $n=3,,(a_1,a_2,a_3)=(0,4,3)$ and $x=(5,0,1)$.
$endgroup$
Random counterexample: $Phi(x)=-288$ when $n=3,,(a_1,a_2,a_3)=(0,4,3)$ and $x=(5,0,1)$.
answered Jan 16 at 17:04
user1551user1551
74.7k566129
74.7k566129
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