Prove that the functional is non-negative for $x_igeq 0$












3












$begingroup$


Consider the following functional $Phi:mathbb R^ntomathbb R $:
$$
Phi(x)=sum_{i=1}^{n-1}(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1).
$$
The computer experiments show that it is non-negative for all $x_igeq 0$. I need to prove this. Note that we have both $Phi(x)=0$ and $nabla Phi(x)=0$ for all $x$ with equal coordinates. The proof should be simple, but I can't manage to find it. Any ideas?






inequality







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$endgroup$








  • 1




    $begingroup$
    $Phi$ is not nonnegative on $mathbb R^n$. E.g. when $x_1=x_2=cdots=x_{n-1}=1$, $Phi(x)=6(n-1)(x_n-1)^2(x_n+1)$, which is negative when $x_n+1<0$.
    $endgroup$
    – user1551
    Jan 11 at 20:00










  • $begingroup$
    @user1551 thank you very much, you are right. I corrected the question.
    $endgroup$
    – Nasa Momdele
    Jan 11 at 20:19
















3












$begingroup$


Consider the following functional $Phi:mathbb R^ntomathbb R $:
$$
Phi(x)=sum_{i=1}^{n-1}(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1).
$$
The computer experiments show that it is non-negative for all $x_igeq 0$. I need to prove this. Note that we have both $Phi(x)=0$ and $nabla Phi(x)=0$ for all $x$ with equal coordinates. The proof should be simple, but I can't manage to find it. Any ideas?






inequality







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $Phi$ is not nonnegative on $mathbb R^n$. E.g. when $x_1=x_2=cdots=x_{n-1}=1$, $Phi(x)=6(n-1)(x_n-1)^2(x_n+1)$, which is negative when $x_n+1<0$.
    $endgroup$
    – user1551
    Jan 11 at 20:00










  • $begingroup$
    @user1551 thank you very much, you are right. I corrected the question.
    $endgroup$
    – Nasa Momdele
    Jan 11 at 20:19














3












3








3


0



$begingroup$


Consider the following functional $Phi:mathbb R^ntomathbb R $:
$$
Phi(x)=sum_{i=1}^{n-1}(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1).
$$
The computer experiments show that it is non-negative for all $x_igeq 0$. I need to prove this. Note that we have both $Phi(x)=0$ and $nabla Phi(x)=0$ for all $x$ with equal coordinates. The proof should be simple, but I can't manage to find it. Any ideas?






inequality







share|cite|improve this question











$endgroup$




Consider the following functional $Phi:mathbb R^ntomathbb R $:
$$
Phi(x)=sum_{i=1}^{n-1}(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1).
$$
The computer experiments show that it is non-negative for all $x_igeq 0$. I need to prove this. Note that we have both $Phi(x)=0$ and $nabla Phi(x)=0$ for all $x$ with equal coordinates. The proof should be simple, but I can't manage to find it. Any ideas?






inequality




inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 20:19







Nasa Momdele

















asked Jan 10 at 22:16









Nasa MomdeleNasa Momdele

254




254








  • 1




    $begingroup$
    $Phi$ is not nonnegative on $mathbb R^n$. E.g. when $x_1=x_2=cdots=x_{n-1}=1$, $Phi(x)=6(n-1)(x_n-1)^2(x_n+1)$, which is negative when $x_n+1<0$.
    $endgroup$
    – user1551
    Jan 11 at 20:00










  • $begingroup$
    @user1551 thank you very much, you are right. I corrected the question.
    $endgroup$
    – Nasa Momdele
    Jan 11 at 20:19














  • 1




    $begingroup$
    $Phi$ is not nonnegative on $mathbb R^n$. E.g. when $x_1=x_2=cdots=x_{n-1}=1$, $Phi(x)=6(n-1)(x_n-1)^2(x_n+1)$, which is negative when $x_n+1<0$.
    $endgroup$
    – user1551
    Jan 11 at 20:00










  • $begingroup$
    @user1551 thank you very much, you are right. I corrected the question.
    $endgroup$
    – Nasa Momdele
    Jan 11 at 20:19








1




1




$begingroup$
$Phi$ is not nonnegative on $mathbb R^n$. E.g. when $x_1=x_2=cdots=x_{n-1}=1$, $Phi(x)=6(n-1)(x_n-1)^2(x_n+1)$, which is negative when $x_n+1<0$.
$endgroup$
– user1551
Jan 11 at 20:00




$begingroup$
$Phi$ is not nonnegative on $mathbb R^n$. E.g. when $x_1=x_2=cdots=x_{n-1}=1$, $Phi(x)=6(n-1)(x_n-1)^2(x_n+1)$, which is negative when $x_n+1<0$.
$endgroup$
– user1551
Jan 11 at 20:00












$begingroup$
@user1551 thank you very much, you are right. I corrected the question.
$endgroup$
– Nasa Momdele
Jan 11 at 20:19




$begingroup$
@user1551 thank you very much, you are right. I corrected the question.
$endgroup$
– Nasa Momdele
Jan 11 at 20:19










2 Answers
2






active

oldest

votes


















1












$begingroup$

$Phi$ is not always nonnegative. Let $f(x_i)=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)$. When $x_1=7,,x_2=x_3=cdots=x_{n-1}=1$ and $x_n=0$, we have, for $2le ile n-1$,
begin{aligned}
f(x_i)
&=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)\
&=2(1)^2left(2(1+1+0)+0-7right)\
&=-6.
end{aligned}
Therefore $Phi(x)=sum_{i=1}^{n-1}f(x_i)=f(x_1)-6(n-2)$ is negative when $n$ is sufficiently large.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you please take a look at a similar inequality? math.stackexchange.com/q/3075042/633591
    $endgroup$
    – Nasa Momdele
    Jan 15 at 22:16





















1












$begingroup$

Not an answer, merely recording a partial result, found trying to find the least $n$ with a $Phi_n(x) < 0$.



begin{align*}
Phi_{78}(frac{13018}{3}, 1626, dots, 1626,0) = frac{-868,706,947,328}{81}
end{align*}
In fact, $Phi_{78}left( frac{25979 b}{9736}, b, dots, b, 0 right)$ has constant negative fourth derivative and has negative first, second, and third derivatives as soon as $b > 1200.487dots$, so $Phi_{78}$ decreases like $Theta(-b^4)$ starting there. (More on $Theta$.)






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    1












    $begingroup$

    $Phi$ is not always nonnegative. Let $f(x_i)=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)$. When $x_1=7,,x_2=x_3=cdots=x_{n-1}=1$ and $x_n=0$, we have, for $2le ile n-1$,
    begin{aligned}
    f(x_i)
    &=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)\
    &=2(1)^2left(2(1+1+0)+0-7right)\
    &=-6.
    end{aligned}
    Therefore $Phi(x)=sum_{i=1}^{n-1}f(x_i)=f(x_1)-6(n-2)$ is negative when $n$ is sufficiently large.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Could you please take a look at a similar inequality? math.stackexchange.com/q/3075042/633591
      $endgroup$
      – Nasa Momdele
      Jan 15 at 22:16


















    1












    $begingroup$

    $Phi$ is not always nonnegative. Let $f(x_i)=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)$. When $x_1=7,,x_2=x_3=cdots=x_{n-1}=1$ and $x_n=0$, we have, for $2le ile n-1$,
    begin{aligned}
    f(x_i)
    &=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)\
    &=2(1)^2left(2(1+1+0)+0-7right)\
    &=-6.
    end{aligned}
    Therefore $Phi(x)=sum_{i=1}^{n-1}f(x_i)=f(x_1)-6(n-2)$ is negative when $n$ is sufficiently large.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Could you please take a look at a similar inequality? math.stackexchange.com/q/3075042/633591
      $endgroup$
      – Nasa Momdele
      Jan 15 at 22:16
















    1












    1








    1





    $begingroup$

    $Phi$ is not always nonnegative. Let $f(x_i)=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)$. When $x_1=7,,x_2=x_3=cdots=x_{n-1}=1$ and $x_n=0$, we have, for $2le ile n-1$,
    begin{aligned}
    f(x_i)
    &=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)\
    &=2(1)^2left(2(1+1+0)+0-7right)\
    &=-6.
    end{aligned}
    Therefore $Phi(x)=sum_{i=1}^{n-1}f(x_i)=f(x_1)-6(n-2)$ is negative when $n$ is sufficiently large.






    share|cite|improve this answer









    $endgroup$



    $Phi$ is not always nonnegative. Let $f(x_i)=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)$. When $x_1=7,,x_2=x_3=cdots=x_{n-1}=1$ and $x_n=0$, we have, for $2le ile n-1$,
    begin{aligned}
    f(x_i)
    &=(1+x_i)(x_i-x_n)^2(2(1+x_i+x_n)+x_i x_n-x_1)\
    &=2(1)^2left(2(1+1+0)+0-7right)\
    &=-6.
    end{aligned}
    Therefore $Phi(x)=sum_{i=1}^{n-1}f(x_i)=f(x_1)-6(n-2)$ is negative when $n$ is sufficiently large.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 11 at 20:34









    user1551user1551

    74.3k566129




    74.3k566129












    • $begingroup$
      Could you please take a look at a similar inequality? math.stackexchange.com/q/3075042/633591
      $endgroup$
      – Nasa Momdele
      Jan 15 at 22:16




















    • $begingroup$
      Could you please take a look at a similar inequality? math.stackexchange.com/q/3075042/633591
      $endgroup$
      – Nasa Momdele
      Jan 15 at 22:16


















    $begingroup$
    Could you please take a look at a similar inequality? math.stackexchange.com/q/3075042/633591
    $endgroup$
    – Nasa Momdele
    Jan 15 at 22:16






    $begingroup$
    Could you please take a look at a similar inequality? math.stackexchange.com/q/3075042/633591
    $endgroup$
    – Nasa Momdele
    Jan 15 at 22:16













    1












    $begingroup$

    Not an answer, merely recording a partial result, found trying to find the least $n$ with a $Phi_n(x) < 0$.



    begin{align*}
    Phi_{78}(frac{13018}{3}, 1626, dots, 1626,0) = frac{-868,706,947,328}{81}
    end{align*}
    In fact, $Phi_{78}left( frac{25979 b}{9736}, b, dots, b, 0 right)$ has constant negative fourth derivative and has negative first, second, and third derivatives as soon as $b > 1200.487dots$, so $Phi_{78}$ decreases like $Theta(-b^4)$ starting there. (More on $Theta$.)






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Not an answer, merely recording a partial result, found trying to find the least $n$ with a $Phi_n(x) < 0$.



      begin{align*}
      Phi_{78}(frac{13018}{3}, 1626, dots, 1626,0) = frac{-868,706,947,328}{81}
      end{align*}
      In fact, $Phi_{78}left( frac{25979 b}{9736}, b, dots, b, 0 right)$ has constant negative fourth derivative and has negative first, second, and third derivatives as soon as $b > 1200.487dots$, so $Phi_{78}$ decreases like $Theta(-b^4)$ starting there. (More on $Theta$.)






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Not an answer, merely recording a partial result, found trying to find the least $n$ with a $Phi_n(x) < 0$.



        begin{align*}
        Phi_{78}(frac{13018}{3}, 1626, dots, 1626,0) = frac{-868,706,947,328}{81}
        end{align*}
        In fact, $Phi_{78}left( frac{25979 b}{9736}, b, dots, b, 0 right)$ has constant negative fourth derivative and has negative first, second, and third derivatives as soon as $b > 1200.487dots$, so $Phi_{78}$ decreases like $Theta(-b^4)$ starting there. (More on $Theta$.)






        share|cite|improve this answer









        $endgroup$



        Not an answer, merely recording a partial result, found trying to find the least $n$ with a $Phi_n(x) < 0$.



        begin{align*}
        Phi_{78}(frac{13018}{3}, 1626, dots, 1626,0) = frac{-868,706,947,328}{81}
        end{align*}
        In fact, $Phi_{78}left( frac{25979 b}{9736}, b, dots, b, 0 right)$ has constant negative fourth derivative and has negative first, second, and third derivatives as soon as $b > 1200.487dots$, so $Phi_{78}$ decreases like $Theta(-b^4)$ starting there. (More on $Theta$.)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 11 at 23:04









        Eric TowersEric Towers

        33.9k22370




        33.9k22370






























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