Prove that the functional is non-negative for $x_i\geq 0$

Question

Consider the following functional $\Phi:\mathbb R^n\to\mathbb 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 $x_i\geq 0$. 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?

Answer 1

$\Phi$ 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 $2\le i\le n-1$, \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)^2\left(2(1+1+0)+0-7\right)\\ &=-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.

Answer 2

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 $b > 1200.487\dots$, so $\Phi_{78}$ decreases like $\Theta(-b^4)$ starting there. (More on $\Theta$.)