let $a,b,c,x,y,z>0$ such $x+y+z=a+b+c,abc=xyz$,and $a>\max\{x,y,z\}$,


I conjecture $$a^n+b^n+c^n\ge x^n+y^n+z^n,\forall n\in N^{+}$$


Maybe this kind of thing has been studied, like I found something relevant, but I didn't find the same one.[Schur convexity and Schur multiplicative convexity for a class of symmetric function]


I found sometime maybe [The sum of squared logarithms conjecture](https://mathoverflow.net/questions/207845/the-sum-of-squared-logarithms-conjecture)



Conjecture: $a^n+b^n+c^n\ge x^n+y^n+z^n$
A triple $(a,b,c)$ with $a+b+c=s$, $abc=p$, and $ab+bc+ca=t$ is the triple of roots of $X^3-sX^2+tX-p=0$, i.e., of $-X^2+sX+p/X=t$. Fix $s$ and $p$; the left-hand part has just two local extrema on the right semiaxis. Let $a(t)>b(t)>c(t)$ be the three roots (defined for all values of $t$ when they exist and are sdistinct). Looking at the graph, one can easily see that $a(t)$ and $c(t)$ decrease as $t$ grows, while $b(t)$ grows as well.


Now it suffices to prove that $(a^n+b^n+c^n)'=n(a^{n-1}a'+b^{n-1}b'+c^{n-1}c')\leq 0$ (the derivative is taken with respect to $t$). We know that $a'+b'+c'=a'bc+b'ca+c'ab=0$, hence $a'=\lambda a(b-c)$, $b'=\lambda b(c-a)$, and $c'=\lambda c(a-b)$, where $\lambda<0$ as $a'<0$. Hence the required inequality reads $a^n(b-c)+b^n(c-a)+c^n(a-b)\geq0$, or
$$
 a(b-c)(a^{n-1}-b^{n-1})\geq c(a-b)(b^{n-1}-c^{n-1}),
$$
which follows from
$$
 \frac{a^{n-1}-b^{n-1}}{a-b}\geq\frac{b^{n-1}-c^{n-1}}{b-c}.
$$
The last inequality holds for all (not neccessarily integer) $n\geq 2$, e.g., by Lagrange's mean value theorem.


