- #1
cloud18
- 8
- 0
If a1 >= a2 >= ... >= a_n and b1 >= b2 >= ... >= b_n, prove that:
( Sum(a_k from k = 1 to n) )*( Sum(b_k from k = 1 to n) ) <= n*Sum(a_k*b_k
from k = 1 to n).
Hint: Sum( (a_k - a_j)*(b_k - b_j) s.t. 1 <= j <= k <= n ) >= 0
This is what I have so far:
0 <= Sum( (a_k - a_j)*(b_k - b_j) s.t. 1 <= j <= k <= n )
0 <= Sum( a_k*b_k - a_k*b_j - a_j*b_k + a_j*b_j s.t. 1 <= j <= k <= n )
0 <= Sum( a_k*b_k from k = 1 to n ) - Sum( a_k*b_j s.t. 1 <= j <= k <=
n ) - Sum( a_j*b_k s.t. 1 <= j <= k <= n ) + Sum( a_j*b_j from j = 1 to
n )
0 <= 2*Sum( a_k*b_k from k = 1 to n ) - Sum( a_k*b_j s.t. 1 <= j <= k <=
n ) - Sum( a_j*b_k s.t. 1 <= j <= k <= n )
I hope that is correct thus far, but do not know what to do next.
( Sum(a_k from k = 1 to n) )*( Sum(b_k from k = 1 to n) ) <= n*Sum(a_k*b_k
from k = 1 to n).
Hint: Sum( (a_k - a_j)*(b_k - b_j) s.t. 1 <= j <= k <= n ) >= 0
This is what I have so far:
0 <= Sum( (a_k - a_j)*(b_k - b_j) s.t. 1 <= j <= k <= n )
0 <= Sum( a_k*b_k - a_k*b_j - a_j*b_k + a_j*b_j s.t. 1 <= j <= k <= n )
0 <= Sum( a_k*b_k from k = 1 to n ) - Sum( a_k*b_j s.t. 1 <= j <= k <=
n ) - Sum( a_j*b_k s.t. 1 <= j <= k <= n ) + Sum( a_j*b_j from j = 1 to
n )
0 <= 2*Sum( a_k*b_k from k = 1 to n ) - Sum( a_k*b_j s.t. 1 <= j <= k <=
n ) - Sum( a_j*b_k s.t. 1 <= j <= k <= n )
I hope that is correct thus far, but do not know what to do next.