Find the minimun value of $\sum_{i=1}^{10} \sum_{j=1}^{10} \sum_{k=1}^{10}|k(x+y-10i)(3x-6y-36j)(19x+95y-95k)|$ , where $x,y$ are integers.
Problem
Source: China mathematical olympiad 1995 problem3
Tags: inequalities, inequalities unsolved, Sums and Products
22.02.2015 18:51
Does anyone have a solution to this complicated-looking inequality? Thanks!
24.01.2016 17:06
any solution?
28.09.2016 11:37
MathPanda1 wrote: Does anyone have a solution to this complicated-looking inequality? Thanks! So you know that this inequality is only just complicated-looking ,why don't you try and find it quite easy? My solution It is quite obvious that $\sum_{i=1}^{10} \sum_{j=1}^{10} \sum_{k=1}^{10}|k(x+y-10i)(3x-6y-36j)(19x+95y-95k)|=\sum_{i=1}^{10}|(x+y-10i)|\sum_{j=1}^{10}|(3x-6y-36j)|\sum_{k=1}^{10}|k(19x+95y-95k)|$ so it is sufficent to find the minimun value of $\sum_{i=1}^{10}|(x+y-10i)|$,$\sum_{j=1}^{10}|(3x-6y-36j)|$and$\sum_{k=1}^{10}|k(19x+95y-95k)|$ respectively but this part is also very easy ,we only need to use some property about abs function ,em now we can immediately get that $\sum_{i=1}^{10}|(x+y-10i)|\geq{250}$ when $50\le{x+y}\le{60}$ the equality holds and $\sum_{j=1}^{10}|(3x-6y-36j)|\geq{900}$ when $60\le{x-2y}\le{72}$ the equality holds and finally $\sum_{k=1}^{10}|k(19x+95y-95k)|\geq{95*112}$ when $x+5y=35 $the equality holds ,now when we take $(x,y)=(60,-5)$ then the all equalities hold and we can get the minimun value is $2394000000$