Question: Let $x,y,z\in\mathbb{R}$. How to eliminate quantifier $t$ in “For all $t\in\mathbb{N}$, $x^t(x-y)(x-z)+y^t(y-z)(y-x)+z^t(z-x)(z-y)\ge0$” so that we have a statement about only $x,y$ and $z$?
My attempt:
Firstly, take $t=1$ and consider the case "all $x,y,z$ are non-positive"
Suppose $x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y)\ge0$ and $x\le0,y\le0,z\le0$
Apply Schur's inequality to $-x,-y,-z$, we have$$x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y)\le0$$therefore$$x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y)=0$$it reduces to the case for equality of Schur's inequality.
Therefore in the case "all $x,y,z$ are non-positive" the statement is equivalent to$$[x=y=z\le0]\vee[x=y<0\wedge z=0]\vee[x=z<0\wedge y=0]\vee[y=z<0\wedge x=0]$$
Now assume one of $x,y,z$ is positive. Wlog $z=\max(x,y,z)>0$.
Since the inequality is homogeneous, scaling $x,y,z$ by a factor $1/z$ we may assume $z=1$. The statement becomes
$$x^t(x-y)(x-1)+y^t(y-1)(y-x)+(1-x)(1-y)\ge0\quad\forall t\in\mathbb{N}$$
Claim 1. When $t$ is even, the above inequality is true for all $x,y\in\mathbb{R}$.
I separated this to a new question.
So we only consider odd numbers $2t-1$,
Claim 2. The limit of the region$$\mathcal{R}_t=\{(x,y)\in\mathbb{R}^2\mid x^{2t-1}(x-y)(x-1)+y^{2t-1}(y-1)(y-x)+(1-x)(1-y)\ge0\}$$ as $t\to\infty$ is the region $\{(x,y)\in\mathbb{R}^2\mid\max(1-\max(|x|,|y|),x+y)\ge0\vee y=x\}$
I separated this to a new question.
Claim 3. $$\mathcal{R}_1\cap\mathcal{R}_\infty\subseteq\mathcal{R}_t\quad\forall t$$
Proof: I don't know how to prove.
For $t=3$ WolframAlpha says it is true.
For $t=5$ WolframAlpha says it is true.
For $t=7$ WolframAlpha says it is true.
How do you prove Claim 3?
If Claim 1,2,3 are proved, we will get the conclusion:
“For all $t\in\mathbb{N}$, $x^t(x-y)(x-z)+y^t(y-z)(y-x)+z^t(z-x)(z-y)\ge0$” is equivalent to: $$[x=y=z\le0]\vee[x=y<0\wedge z=0]\vee[x=z<0\wedge y=0]\vee[y=z<0\wedge x=0]\vee \mathbf P_x\vee\mathbf P_y\vee\mathbf P_z $$where$$\mathbf P_z:=[z>0]\wedge[x(x-y)(x-z)+y(y-z)(y-x)+(z-x)(z-y)\ge0]\wedge[\max(z-\max(|x|,|y|),x+y)\ge0\vee y=x]$$and define $\mathbf P_x,\mathbf P_y$ similarly.
