Given integer $m\geq2$,$x_1,...,x_m$ are non-negative real numbers,prove that:$$(m-1)^{m-1}(x_1^m+...+x_m^m)\geq(x_1+...+x_m)^m-m^mx_1...x_m$$and please find out when the equality holds.
Problem
Source: 2017 China TST 4 Problem 5
Tags: inequalities, TST
guangzhou-2015
22.03.2017 19:12
nice problem
angiland
22.03.2017 19:50
Fix the sum to be 1 and apply Lagrange multipliers.
anantmudgal09
22.03.2017 21:20
For $m=2$ the inequality is obvious. Let $m \ge 3$ and WLOG, $x_1+\dots+x_m=1.$ Our objective is to minimise the function $$f(x_1, x_2, \dots ,x_m)=(m-1)^{m-1}\left(x_1^m+\dots+x_m^m\right)+m^m\cdot x_1x_2\dots x_m,$$subject to the constraint $g(x_1, \dots, x_m)=x_1+\dots+x_m=1.$ Note that $\nabla g=(1,\dots, 1) \ne 0$ and $f, g$ are continuous functions with continuous partial derivatives. Let $U=[0,1]^m$ and note that the set $$\{x: x \in U, g(x)=1\}$$is a closed and bounded subset of $\mathbb{R}^m$ hence it is compact. Therefore, $f$ attains a minima over this set (by Weistrass' Theorem).
For a point on the boundary, we get $0 \in \{x_1, \dots, x_m\}$ where the result is obvious by applying the Power Mean Inequality. Otherwise, set $x=(x_1, \dots, x_m)$ and note that $$\nabla f =\lambda \nabla g,$$for some constant $\lambda$. Fix $t=m^mx_1x_2\dots x_m$ and $s=(m-1)^{m-1}$ and observe that $$\lambda=\frac{t}{x}+sx^{m-1},$$where $x \in \{x_1, \dots, x_m\}$ is a real number. Note that for $x \ne y$ we have $$\lambda=\frac{t}{x}+sx^{m-1}=\frac{t}{y}+sy^{m-1} \Longrightarrow \lambda=s\frac{y^m-x^m}{y-x}.$$Moreover, $$\frac{t}{xy}=s\frac{y^{m-1}-x^{m-1}}{y-x}=\lambda \frac{y^{m-1}-x^{m-1}}{y^m-x^m} \Longrightarrow \frac{t}{\lambda}=\frac{xy^m-yx^m}{y^m-x^m}.$$Evidently, for $z>0$ repeating the same argument gives $$\frac{t}{\lambda}=\frac{\left(xy^m-yx^m\right)-\left(xz^m-zx^m\right)}{\left(y^m-x^m\right)-\left(z^m-x^m\right)}=\frac{zy^m-yz^m}{y^m-z^m} \Longrightarrow \sum_{x,y,z} x^m(y-z)=0.$$Finally, we remark that by Schur's Inequality, the last result holds if and only if $\{z, 0\}=\{x, y\}.$ Evidently, we only need to check that the result holds when all variables are equal, which is obvious.
The only equality cases are $(0,1,\dots, 1)$ and it's cyclic permutations. $\square$
This was a neat problem, in the sense that we ended up using Schur to finish the LM solution.
MathPanda1
23.03.2017 01:15
Is there possibly a solution without using Lagrange Multipliers? And is Lagrange Multipliers required in e.g. IMO/USAMO/CMO(Canada)/APMO? Thank you very much!
Jettofaiyafukushireikan
08.04.2017 05:35
HuangZhen wrote: Given integer $m\geq2$,$x_1,...,x_m$ are non-negative real numbers,prove that:$$(m-1)^{m-1}(x_1^m+...+x_m^m)\geq(x_1+...+x_m)^m-m^mx_1...x_m$$and please find out when the equality holds. The problem supposed to be prove that:$$(m-1)^{m-1}(x_1^m+...+x_m^m-mx_1x_2...x_m)\geq(x_1+...+x_m)^m-m^mx_1...x_m$$
liekkas
09.06.2017 06:06
Any clean solutions to it?
CQYIMO42
09.06.2017 06:15
liekkas wrote: Any clean solutions to it? I have a solution, I can write it down later (edit: nvm my solution is pretty long, though I can still post it if you want to read :]]])
MathLife00
30.10.2017 19:26
What is Lagrange Multipliers? I can't understand the solution...
me9hanics
30.10.2017 19:41
And are Lagrange Multipliers worth to learn for Olympiads?
Justanaccount
24.05.2021 10:40
I would like to point out that anantmudgal09 's solution is wrong, because it only works when x,y,z are pairwise dinstict. You can't divide x-z and y-z when z is equal to x or y.
mihaig
08.06.2021 09:13
Jettofaiyafukushireikan wrote: The problem supposed to be prove that:$$(m-1)^{m-1}(x_1^m+...+x_m^m-mx_1x_2...x_m)\geq(x_1+...+x_m)^m-m^mx_1...x_m$$ See also here https://artofproblemsolving.com/community/c6t243f6h2584785_shleifer_like_inequality
Rhapsodies_pro
03.03.2024 17:22
liekkas wrote: Any clean solution to it? One may make use of the so-called Equal Variables Method for non-negative variables (cf. 幾道集訓隊試題的解與評析——孔繁浩（東北育才學校）.PDF). Anyway, it has been additionally proved
《走向IMO：數學奧林匹克試題集錦（2017）》第138～140頁
that \[{\left(1-\frac1m\right)}^{m-1}{\left(\frac{\operatorname{\textsl p_\mathit k\!}{\left(x_1,x_2\dotsc,x_m\right)}}m-\frac{\operatorname{\textsl e_\mathit k\!}{\left(x_1,x_2\dotsc,x_m\right)}}{\binom mk}\right)}\geqq{\left(\frac{\operatorname{\textsl p_\mathrm 1\!}{\left(x_1,x_2\dotsc,x_m\right)}}m\right)}^{\!k}-\frac{\operatorname{\textsl e_\mathit k\!}{\left(x_1,x_2\dotsc,x_m\right)}}{{m\choose k}}\textnormal,\]where \(k\) is a natural number satisfying \(k\leqq m\).