Find all integer $n$ such that the following property holds: for any positive real numbers $a,b,c,x,y,z$, with $max(a,b,c,x,y,z)=a$ , $a+b+c=x+y+z$ and $abc=xyz$, the inequality $$a^n+b^n+c^n \ge x^n+y^n+z^n$$holds.
Problem
Source: 2019 China TST Test 4 P5
Tags: inequalities, China TST, China
06.04.2019 19:21
Solving (partially) this outstanding problems,I created two problems. I hope the readers we'll like them and I also hope that they shall help the reader to finish the main task.
Attachments:

07.04.2019 00:32
17.04.2019 13:52
Pathological wrote:
Any motivation for counter example and proof?
15.05.2019 17:00
answer: Every $n\ge 0$ works. For $n<0$ , $$(a, b, c) = (2, 3, 4), (x, y, z) = (3.5, \frac{5.5 + \sqrt{5.5^2 - 4*48/7}}{2}, \frac{5.5 - \sqrt{5.5^2 - 4*48/7}}{2})$$Is a counterexample @thanksabove Now $n\ge 0$ . lemma1: If $max(a,b,c)\ge max(x,y,z)$ then $ab+bc+ca\le xy+yz+zx$ It follows from the fact that $(a-x)(a-y)(a-z)\ge (a-a)(a-b)(a-c)$ , $a(xy+yz+zx)\ge a(ab+bc+ca)$ This would be obvious anyway later on the solution, it also proves case $n=2$. We are going to prove the following, using calculus: Fix $a+b+c=s$ , $abc=1$ and denote $ab+bc+ca=p$. $a^n+b^n+c^n$ decreases as $p$ increases. We investigate function $$f(x)=x^2-sx-\frac{1}{x}$$for $x>0$. Note that $f(x)+p=\frac{(x-a)(x-b)(x-c)}{x}$. $f'(x)=2x+\frac{1}{x^2}-s$. As $s>3$ , $f'(1) <0$ figure out that $f'(r_1)=f'(r_2)=0$ for which $r_1<1<r_2<s $ , $f(r_1),f(r_2)<0$. and at this point we come to describe the graph of $f$. $f(x)$ tends to $- \infty$ as $x$ tends to $0$ $f(x)$ increases in $(0,r_1)$ , decreases in $(r_1,r_2)$ , increases in $(r_2,+\infty)$ The gragh of $f$ is under $y=0$ for $x\in(0,s)$. Given the form of $f$ the line $y=-p$ intersects $y=f(x)$ in three points $c<r_1<b<r_2<a$ as long as $-p\in (f(r_2),f(r_1))$ Now we investigate $$g(x)=f(x^{1/n})=x^{2/n}-sx^{1/n}-x^{-1/n}$$. $g$ has the same increasing,decreasing,increasing form and line $y=-p$ intersects $g$ in three different points $c^n<r^n_1<b^n<r^n_2<a^n$ as long as $-p$ is between $g(r^n_2)$ and $g(r^n_1)$. Denote $a(p)=a^n=g^{-1}(-p)$ in interval $(r^n_2,+\infty)$. $g^{-1}$ is differentiable in this interval so $\frac{da(p)}{dp}=\frac{dg^{-1}(-p)}{dp}$ $a'(p)=\frac{-1}{g'(a^n)}$ , $g'(x)=\frac{1}{xn}(2x^{2/n}+x^{-1/n}-sx^{1/n})$ , $g'(a^n)=\frac{1}{na^n}(2a^2+1/a-(a+b+c)a)=\frac{1}{na^n}(a^2+bc-ab-ac)=\frac{1}{na^n}(a-b)(a-c)$. We actually need $a'(p)+b'(p)+c'(p)<0$ (define $b(p),c(p)$ in a similar way) which reduces to $$\frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)}>0$$as $c<b<a$ , $(a-b)(b-c)(a-c)>0$ multiply $$a^n(b-c)+b^n(c-a)+c^n(a-b)>0$$$$(a^n-b^n)(b-c)>(b^n-c^n)(a-b)$$Which is obvious , thus $a(p)+b(p)+c(p)=a^n+b^n+c^n$ decreases as $p$ increases.
17.05.2019 12:56
If letting $n$ be real number, it changes $\cdots\cdots$ When $0 < n < 1$, it does not work. For example, $n= \frac{1}{2}$, $a = 6, b=3, c=2$ and $x = 5, y = 3 + \frac{3}{\sqrt{5}}, z= 3 - \frac{3}{\sqrt{5}}$. However, $\sqrt{a}+\sqrt{b}+\sqrt{c} < \sqrt{x} + \sqrt{y} + \sqrt{z}$.
17.05.2019 15:12
It is similar to my old post. https://artofproblemsolving.com/community/u473756h1833674p12287396 Let $a, b, c, u, v, w$ be non-negative real numbers satisfying $$a+ b+ c = u+v+w, \quad ab + bc + ca = uv + vw + wu, \quad abc \ge uvw.$$Let $f(u)$ be a function with $f'''(u)\ge 0, \ \forall u > 0$. Prove that $f(a)+f(b)+f(c) \ge f(u) + f(v) + f(w)$.
17.05.2019 23:47
Quote: If letting $n$ be real number, it changes $\cdots\cdots$ When $0 < n < 1$, it does not work. For example, $n= \frac{1}{2}$, $a = 6, b=3, c=2$ and $x = 5, y = 3 + \frac{3}{\sqrt{5}}, z= 3 - \frac{3}{\sqrt{5}}$. However, $\sqrt{a}+\sqrt{b}+\sqrt{c} < \sqrt{x} + \sqrt{y} + \sqrt{z}$.] My solution proves the inequality all real $n\ge2$ . I also prove the fact $h(a)+h(b)+h(c)\ge h(x)+h(y)+h(z)$ for three times differentiable $h:[0,+\infty )\to [0,+\infty )$ ,. and $h"(x),h'''(x)>0$. Modify the above arguments so, we basically calculate (I'm silly the definition of $g$ is unnecessery)$$\frac{\partial a}{\partial p}$$but $a=f^{-1}(-p)$ so $$\frac{df^{-1}(-p)}{dp}=-\frac{1}{f'(a)}=-\frac{1}{2a+1/a^2-s}=-\frac{a}{(a-b)(a-c)}$$$$\frac{\partial h(a)}{\partial p}=-\frac{ah'(a)}{(a-b)(a-c)}$$So we have to prove that $(a>b>c)$ $$\frac{ah'(a)}{(a-b)(a-c)}+\frac{bh'(b)}{(b-c)(b-a)}+\frac{ch'(c)}{(c-a)(c-b)}>0$$$$ah'(a)(b-c)+bh'(b)(c-a)+ch'(c)(a-b) >0$$$$\frac{ah'(a)-bh'(b)}{a-b}>\frac{bh'(b)-ch'(c)}{b-c}$$which is trivial by mean value theorem . So here is your problem : Function $f:(0,+\infty )\to (0,+\infty ) $ is three times differentiable and $f''(x),f'''(x)\ge 0$. Positive reals $a,b,c,x,y,z $ are such that $max(a,b,c)\ge max(x,y,z)$ and $a+b+c=x+y+z$ , $abc=xyz$ . Prove that $f(a)+f(b)+f(c)\ge f(x)+f(y)+f(z)$.
18.05.2019 03:38
Dear mela_20-15, nice solution. Yesterday, I used my method in my old post to get the condition $f''(x) > 0$. https://artofproblemsolving.com/community/c6h1833674p12287396 The main part of my solution is the following: (the idea is similar to the above link) Lemma 1: $F'(q) \le 0$ for $q \in (q_1, q_2)$. Proof of Lemma 1: When $q \in (q_1, q_2)$, $t^3-pt^2+qt-r=0$ has three distinct positive real roots $u, v, w$. WLOG, assume $u > v > w$. Using implicit differentiation for the system of equations $p=u+v+w, \ q = uv+vw+wu, \ r = uvw$, we have $\frac{\partial u}{\partial q} = \frac{-u}{(u-v)(u-w)}, \ \frac{\partial v}{\partial q} = \frac{-v}{(v-u)(v-w)}, \ \frac{\partial w}{\partial q} = \frac{-w}{(w-u)(w-v)}$. Thus, we have $$F'(q) = f'(u)\frac{\partial u}{\partial q} + f'(v) \frac{\partial v}{\partial q} + f'(w) \frac{\partial w}{\partial q} = \frac{-u f'(u)}{(u-v)(u-w)} + \frac{-vf'(v)}{(v-u)(v-w)} + \frac{-wf'(w)}{(w-u)(w-v)}.$$Since $f'''(s)\ge 0, \forall s> 0$, $f'(s)$ is convex on $(0, \infty)$ and hence $\frac{f'(u)-f'(v)}{u-v}\ge \frac{f'(u)-f'(w)}{u-w}$ which results in $f'(u)(v-w) \ge f'(v)(u-w) - f'(w)(u-v)$. Thus, we have $$F'(q) \le \frac{-u (f'(v)(u-w) - f'(w)(u-v)) }{(v-w)(u-v)(u-w)} + \frac{-vf'(v)}{(v-u)(v-w)} + \frac{-wf'(w)}{(w-u)(w-v)} = \frac{-uf'(v)}{(u-v)(v-w)} + \frac{uf'(w)}{(u-w)(v-w)} + \frac{-vf'(v)}{(v-u)(v-w)} + \frac{-wf'(w)}{(w-u)(w-v)} = \frac{-(f'(v)-f'(w))}{v-w}.$$ If $f''(s) > 0$, we have $F'(q) \le 0$.
18.05.2019 12:52
Yeah it is actually really simple, You need $f'(x_1)+x_1f''(x_1)>f'(x_2)+x_2f''(x_2)$ for$ a>x_1>b>x_2>c$ (from the mean value) So you need the derivative $2f''(x)+xf'''(x)>0$.
23.05.2019 14:53
Actually during the test, the problem only said: For positive real numbers $a,b,c,x,y,z$, with $max(a,b,c,x,y,z)=a$ , $a+b+c=x+y+z$ and $abc=xyz$, determine all integer $n$ for which the inequality $$a^n+b^n+c^n \ge x^n+y^n+z^n$$holds. This meant that we needed to prove for all $a,b,c,x,y,z$ and $n< 0$, the inequality doesn't hold, rather than giving a counterexample.
10.06.2019 08:02
im considering whether this problem is related to vasc 's EV theorem to a lesser or greater degree.... (and i believe this problem is presumably dut to hjj)
03.11.2020 05:05
Solved with nukelauncher and Th3Numb3rThr33. The answer is \(n\ge0\); to verify \(n<0\) fail, take \((a,b,c)=(16,3,2)\), \((x,y,z)=(12,8,1)\), where \[\frac1{2^{-n}}+\frac1{3^{-n}}+\frac1{16^{-n}}<1<1+\frac1{8^{-n}}+\frac1{12^{-n}}.\] To prove \(n\ge0\) work, we first prove this lemma: Lemma: In \(X^3-sX^2-tX-p=0\) with positive real roots \(a\), \(b\), \(c\), if we express \(a^n+b^n+c^n\) as a polynomial \(P(s,t,p)\), then all terms have positive coefficient. Proof. Easy induction using Newton sums: if \(U_n=a^n+b^n+c^n\), then \[U_n=sU_{n-1}+tU_{n-2}+pU_{n-3}.\]\(\blacksquare\) Now let \(a\), \(b\), \(c\) be the roots of \(X^3-sX^2-t_1X-p=0\) and let \(x\), \(y\), \(z\) be the roots of \(X^3-sX^2-t_2X-p=0\). I contend: Claim: \(t_1\ge t_2\). Proof. Suppose \(t_1<t_2\). Then \[a^3-sa^2-t_2a-p<a^3-sa^2-t_1a-p=0.\]By the intermediate value theorem, one of the roots of \(X^3-sX^2-t_2X-p\) is greater than \(a\), absurd. \(\blacksquare\) Now the desired conclusion follows by combining the claim and the lemma: we have \[a^n+b^n+c^n=P(s,t_1,p)\ge P(s,t_2,p)=x^n+y^n+z^n.\]
03.11.2020 13:43
Very nice solution, TheUltimate123. Note that Vasc has posted in 1989, in the Romanian journal Gazeta Matematica A, no. 3, the following similar theorem: If $f$ is a three differentiable function on I and $$a_1+a_2+a_3=c_1+c_2+c_3,\ \ \ \ \ \ \ \ a_1^2+a_2^2+a_3^2=c_1^2+c_2^2+c_3^2,$$where $a_i,b_i\in$ I , then there is $x\in$ I such that $$f(a_1)+f(a_2)+f(a_3)-f(c_1)-f(c_2)-f(c_3)=\frac 1{2}(a_1a_2a_3-c_1c_2c_3)f'''(x).$$This means that if $f'''\ge 0$ on I, then $f(a_1)+f(a_2)+f(a_3)-f(c_1)-f(c_2)-f(c_3)\ $ has the same sign as $a_1a_2a_3-c_1c_2c_3$ (which has the sign of $\max\{a_1,a_2,a_3\}-\max\{c_1,c_2,c_3\}).$