The document presents a problem to find the minimum possible value of a summation involving positive numbers a1, a2, ..., an and a permutation i1, i2, ..., in of 1,2,...,n. It is shown that the minimum occurs when the numbers are ordered from smallest to largest and paired in the summation accordingly. Applying a lemma, the minimum value is proven to be n.

1. Problem of the Week No.6
Eduardo Enrique Escamilla Saldaña
March 2 2014.
Monterrey, Nuevo León, México
Let 𝑎1 , 𝑎2 , … , 𝑎 𝑛 be positive numbers. Find the smallest possible value of
𝑛
∑
𝑘=1
𝑎𝑘
𝑎𝑖 𝑘
Where 𝑖1 , 𝑖2 , … , 𝑖 𝑛 is a permutation of 1,2, … , 𝑛.
Lemma: For all 𝑎 𝑖 > 0 𝑎𝑛𝑑 𝑖 ≠ 𝑗:
𝑎𝑗
𝑎𝑖
+ ≥2
𝑎𝑗
𝑎𝑖
Proof:
2
(𝑎 𝑖 − 𝑎 𝑗 ) ≥ 0
∴ 𝑎 𝑖 2 − 2𝑎 𝑖 𝑎 𝑗 + 𝑎 2 ≥ 0
𝑖
∴ 𝑎 𝑖 2 + 𝑎 2 ≥ 2𝑎 𝑖 𝑎 𝑗
𝑖
Dividing both sides by 𝑎 𝑖 𝑎 𝑗 > 0 yields:
𝑎𝑗
𝑎 𝑖 2 + 𝑎2
𝑎𝑖
𝑖
= + ≥2
𝑎 𝑖 𝑎𝑗
𝑎𝑗
𝑎𝑖
As desired.
Without loss of generalization let
𝑎1 ≤ 𝑎2 ≤ ⋯ ≤ 𝑎 𝑛
Equivalently:
1
1
1
≤
≤⋯≤
𝑎𝑛
𝑎 𝑛−1
𝑎1
Then
𝑛
∑
𝑘=1
𝑎𝑘
𝑎𝑖 𝑘
Is minimal when every term in the two inequalities are multiplied pairwise in the same order i.e.

2. 𝐴 = 𝑎1 ∗
1
1
1
+ 𝑎2 ∗
+ ⋯+ 𝑎 𝑛 ∗
𝑎𝑛
𝑎 𝑛−1
𝑎1
For if one changes any two terms in the arrangement say, 𝑎 𝑝 ≤ 𝑎 𝑞 then:
If
𝐴 = 𝑎1 ∗
1
1
1
1
+ ⋯+ 𝑎 𝑝 ∗
+⋯+ 𝑎𝑞 ∗
+ ⋯+ 𝑎 𝑛 ∗
𝑎𝑛
𝑎 𝑛−𝑝+1
𝑎 𝑛−𝑞+1
𝑎1
𝐵 = 𝑎1 ∗
1
1
1
1
+⋯+ 𝑎𝑞 ∗
+ ⋯+ 𝑎 𝑝 ∗
+ ⋯+ 𝑎 𝑛 ∗
𝑎𝑛
𝑎 𝑛−𝑝+1
𝑎 𝑛−𝑞+1
𝑎1
Permuting 𝑎 𝑝 , 𝑎 𝑞 gives
But
𝐵− 𝐴 = 𝑎𝑞∗
1
1
1
1
1
1
+ 𝑎𝑝∗
− 𝑎𝑝∗
− 𝑎𝑞∗
= (𝑎 𝑝 − 𝑎 𝑞 )(
−
)≥0
𝑎 𝑛−𝑝+1
𝑎 𝑛−𝑞+1
𝑎 𝑛−𝑝+1
𝑎 𝑛−𝑞+1
𝑎 𝑛−𝑞+1
𝑎 𝑛−𝑝+1
So the sum 𝐴 is minimal since its status as the smallest value is invariant under any perturbation in the arrangement of its terms.
Therefore the minimal sum is
𝐴 = 𝑎1 ∗
(𝑎1 ∗
1
1
1
+ 𝑎2 ∗
+ ⋯+ 𝑎 𝑛 ∗
𝑎𝑛
𝑎 𝑛−1
𝑎1
1
1
1
1
1
1
1
+ 𝑎 𝑛 ∗ ) + (𝑎2 ∗
+ 𝑎 𝑛−1 ∗ ) + ⋯ + (𝑎 𝑖 ∗
+ 𝑎 𝑛−𝑖+1 ∗ ) + ⋯ + 𝑎 𝑛+1 ∗
,
𝑎𝑛
𝑎1
𝑎 𝑛−1
𝑎2
𝑎 𝑛−𝑖+1
𝑎𝑖
𝑎 𝑛+1
2
𝑛 𝑜𝑑𝑑
2
=
(𝑎1 ∗
{
1
1
1
1
1
1
1
1
+ 𝑎 𝑛 ∗ ) + (𝑎2 ∗
+ 𝑎 𝑛−1 ∗ ) + ⋯ + (𝑎 𝑖 ∗
+ 𝑎 𝑛−𝑖+1 ∗ ) + ⋯ + (𝑎 𝑛 ∗
+ 𝑎 𝑛+1 ∗ ), 𝑒𝑣𝑒𝑛
𝑎𝑛
𝑎1
𝑎 𝑛−1
𝑎2
𝑎 𝑛−𝑖+1
𝑎𝑖
𝑎 𝑛+1
𝑎𝑛
2
2
2
Applying the lemma we have that for 𝑛 𝑜𝑑𝑑 there are exactly
plus the constant 𝑎 𝑛+1 ∗
2
1
𝑎 𝑛+1
𝑛−1
terms with the property that
2
𝑎𝑗
𝑎𝑖
+ ≥2
𝑎𝑗
𝑎𝑖
= 1 implying that for this case
2
𝐴≥2∗
𝑛−1
+1= 𝑛
2
𝑛
Similarly for 𝑛 𝑒𝑣𝑒𝑛 there are exactly terms with the property that
2
𝑎𝑗
𝑎𝑖
+ ≥2
𝑎𝑗
𝑎𝑖
So
𝐴≥2∗
𝑛
= 𝑛
2
Therefore the smallest possible value for
𝑛
∑
𝑘=1
𝑎𝑘
𝑎𝑖 𝑘
2

3. Where 𝑖1 , 𝑖2 , … , 𝑖 𝑛 is a permutation of 1,2, … , 𝑛 is
𝑛
Provided that all the terms are positive.