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
22.03.2017 19:12
nice problem
22.03.2017 19:50
Fix the sum to be 1 and apply Lagrange multipliers.
22.03.2017 21:20
This was a neat problem, in the sense that we ended up using Schur to finish the LM solution.
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!
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:$$(m-1)^{m-1}(x_1^m+...+x_m^m-mx_1x_2...x_m)\geq(x_1+...+x_m)^m-m^mx_1...x_m$$
09.06.2017 06:06
Any clean solutions to it?
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 :]]])
30.10.2017 19:26
What is Lagrange Multipliers? I can't understand the solution...
30.10.2017 19:41
And are Lagrange Multipliers worth to learn for Olympiads?
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.
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
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
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 \(k\leqq m\).