3
$\begingroup$

Let $[m]$ denote the finite set $\{0, 1, 2, \ldots, m\}$. For $n$-tuples $i = (i_1,\ldots, i_n)$ and $j = (j_1,\ldots, j_n)$ with entries in $[m]$, we always have the inequality $$ \max \{i_1, j_1\} + \cdots + \max\{i_n, j_n\} \ge \max\{i_1 + \cdots + i_n, j_1 + \cdots +j_n\}. $$ For how many pairs $(i,j) \in [m]^n \times [m]^n$ does equality hold?

  • 0
    Have you tried some small test cases to see if there is even a pattern? – 2017-01-01
  • 0
    Equality holds if for every $n$, either $i_n\le j_n$ or $i_n\ge j_n$. – 2017-01-01
  • 0
    any counter examples guys? – 2017-01-01

2 Answers 2

1

Equality holds if for every $r$, either $i_r\le j_r$ or $i_r\ge j_r$ where $r∈\{1,2,...n\}$.

For no. of pairs where $i_r\le j_r$, no of ways of picking any two numbers from set $\{0,1,2...m\}$ is $(\frac{(m+1)m}{2}+m+1)^n=(\frac{(m+2)(m+1)}{2})^n$ ways.

Same is the number of cases for $i_r\ge j_r$.

Now the equality case i.e. $i_r=j_r$ is being counted twice above, so it must be subtracted once. The no of ways $i_r=j_r$ is $(m+1)^n$

Therefore the total no. of ways in which equality holds is $\ 2((m+1)(m+1))^n-(m+1)^n$$$=(m+1)^n((\frac{m+2}{2})^n-1)$$

  • 1
    In total there are only $(m+1)^{2n}$ pairs, which is less than $(m + 1)^n(2(m +1)^n - 1)$ is general. – 2017-01-01
  • 0
    @ColinTan Rectified. – 2017-01-01
  • 0
    Thanks for your answer, WiCK3D POiSON. – 2017-01-01
0

This proof may be wrong but correct me if it is:

Assume $(x_i,y_i)$ where $1\leq i\leq n$ are points in the plane.

Then $\max(x_i,y_i)=\frac{|x_i-y_i|}{2}+\frac{x_i+y_i}{2}$

Assume $x_i>y_i$ then $x_i-y_i>0$ so $|x_i-y_i|=x_i-y_i$ and $\frac{|x_i-y_i|}{2}+\frac{x_i+y_i}{2}=\frac{x_i-y_i}{2}+\frac{x_i+y_i}{2}=x_i$

So $\sum\limits_{i=1}^{n}\max(x_i,y_i)\geq \max(\sum\limits_{i=1}^{n}x_i,\sum\limits_{i=1}^{n}y_i)$ reduces to

$$\sum\limits_{i=1}^{n}\frac{|x_i-y_i|}{2}+\sum\limits_{i=1}^{n}\frac{x_i+y_i}{2}\geq \frac{|\sum\limits_{i=1}^{n}x_i-\sum\limits_{i=1}^{n}y_i|}{2}+\frac{\sum\limits_{i=1}^{n}x_i+\sum\limits_{i=1}^{n}y_i}{2}$$

which reduces to

$$\sum\limits_{i=1}^{n}|x_i-y_i|\geq|\sum\limits_{i=1}^{n}x_i-\sum\limits_{i=1}^{n}y_i|$$

letting $z_i=x_i-y_i$ we have

$$\sum\limits_{i=1}^{n}|z_i|\geq|\sum\limits_{i=1}^{n}z_i|$$

Which is true for all real numbers.

The only time equality is held is when all $z_i$ are positive or negative or when $x_i\geq y_i$ is true for all $i$ or $x_i