# Conjecture: $a^n+b^n+c^n\ge x^n+y^n+z^n$

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

• I dont think your conjecture is true. $a^n+b^n+c^n$ can be written in terms of $s=a+b+c$, $p=abc$ and $ab+bc+ca=a(s-a)+p/a$. If you fix $s$ and $p$ you can find $a>x$ with $a(s-a)+p/a=x(s-x)+p/x$ and therefore if you define $b,c, y,z$ s.t. $a+b+c=x+y+z=s$ and $abc=xyz=p$ that $a^n+b^n+c^n=x^n+y^n+z^n$ for all $n$. Now if you perturb $x$ slightly you can make $x^n+y^n+z^n$ larger while $a>x$ is still true – user35593 Apr 3 '19 at 5:21
• Now,I can't any Counterexample – function sug Apr 3 '19 at 5:25
• Maybe I was wrong. It turns out that if you fix some $s, p$ then $ab+bc+ca$ is a monotonically increasing in the maximum $a$. Maybe one can prove somithing similar for $a^n+b^n+c^n$ or one can prove that it is monoton in $ab+bc+ca$. – user35593 Apr 3 '19 at 5:42

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.
Yes, it is true. Without loss of generality, $$a\geq b \geq c$$. Let $$a+b+c=s,\ abc=p$$ then $$b=\frac{(s \ - \ a) \ + \ d}{2}, \ c=\frac{(s \ - \ a) \ - \ d}{2}$$ with $$d=b-c=\sqrt{(s-a)^2- \frac{4p}{a}}$$.
If we consider $$b,c$$ as variables in $$a,s,p$$, we find that $$\begin{eqnarray} \frac{\partial (a^n+b^n+c^n)}{\partial a} &=& na^{n-1}+nb^{n-1}\frac{1}{2}\left(-1+\frac{\partial d}{\partial a}\right)+nc^{n-1}\frac{1}{2}\left(-1-\frac{\partial d}{\partial a}\right)\\ &=& \frac{n}{2}\left(\left(2a^{n-1}-b^{n-1}-c^{n-1}\right)+\frac{\partial d}{\partial a}\left(b^{n-1}-c^{n-1}\right) \right). \end{eqnarray}$$ Hence it is enough to prove the following: $$\frac{\partial d}{\partial a}+\frac{2a^{n-1}-b^{n-1}-c^{n-1}}{b^{n-1}-c^{n-1}}\geq 0$$
We have $$\frac{\partial d}{\partial a}=\frac{a-s+2p/a^2}{d}=\frac{-b-c+2bc/a}{b-c}=\frac{2bc(b^{n-2}+b^{n-3}c+\dots)/a-(b+c)(b^{n-2}+b^{n-3}c+\dots)}{b^{n-1}-c^{n-1}} \$$ Hence, $$\frac{\partial d}{\partial a}+\frac{2a^{n-1}-b^{n-1}-c^{n-1}}{b^{n-1}-c^{n-1}} =\frac{2bc(b^{n-2}+b^{n-3}c+\cdots)/a+2a^{n-1}-(b+c)(b^{n-2}+b^{n-3}c+\dots)-(b^{n-1}+c^{n-1})}{b^{n-1}-c^{n-1}}$$ Taking the derivative with respect to $$a$$ yields $$\frac{\partial^2 d}{\partial a^2} = \frac{2(n-1)a^{n-2} - \frac{2bc}{a^2(b^{n-2} \ + \ b^{n-3}c \ + \ \cdots)}}{b^{n-1} - c^{n-1}}$$ which is greater than or equal to zero. Hence the expression above is minimal if we choose $$a$$ to be minimal, i.e. $$a=b$$. In this case, we have: $$\frac{\partial d}{\partial a}=-1=-\frac{2a^{n-1}-b^{n-1}-c^{n-1}}{b^{n-1}-c^{n-1}}$$.