Long version

7th QEDMO 2010, Problem 8 (a variation on AMM problem #E2353 by J. G. Rau)

Let (a₁, a₂, ..., a_n) be ann-tuple of reals. Let (b₁, b₂, ..., b_n) be ann-tuple of positive reals. We are searching for a permutation π of the set {1,2, ..., n} that minimizes the sum

n

X

k=1

a_π(k)

n

X

i=k

b_π(i) =a_π(1) b_π(1)+b_π(2)+...+b_π(n) +a_π(2) b_π(2)+b_π(3)+...+b_π(n) +...

+a_π(n)b_π(n). Prove that any permutation π that satisfies

a_π(1)

b_π(1) ≤ a_π(2)

b_π(2) ≤...≤ a_π(n) b_π(n) minimizes this sum.

Remark: This problem is slightly stronger than American Mathematical Monthly problem #E2353 by J. G. Rau. My solution below follows R. J. Dickson’s solution in [1].

Solution (according to R. J. Dickson)

We will prove a stronger assertion:

Theorem 1. Let (a₁, a₂, ..., a_n) be ann-tuple of reals. Let (b₁, b₂, ..., b_n) be ann-tuple of positive reals. Letπ be a permutation of the set {1,2, ..., n}.

Then,

n

X

k=1

a_π(k)

n

X

i=k

b_π(i) ≥ X

k∈{1,2,...,n}

a_kb_k+ 1 2

X

(i,k)∈{1,2,...,n}²; i6=k

min{a_kb_i, a_ib_k}. (1)

This inequality (1) becomes an equality if and only if a_π(1)

b_π(1) ≤ a_π(2)

b_π(2) ≤...≤ a_π(n) b_π(n). Proof of Theorem 1. We have

n

X

k=1

a_π(k)

n

X

i=k

b_π(i)

| {z }

=b_π(k)+

n

P

i=k+1

b_π(i)

=

n

X

k=1

a_π(k) b_π(k)+

n

X

i=k+1

b_π(i)

!

=

n

X

k=1

a_π(k)b_π(k)+

n

X

k=1

a_π(k)

n

X

i=k+1

b_π(i)

= X

k∈{1,2,...,n}

akbk+ X

(i,k)∈{1,2,...,n}²; i>k

a_π(k)b_π(i) (2)

(since

n

X

k=1

a_π(k)b_π(k) = X

k∈{1,2,...,n}

a_π(k)b_π(k) = X

k∈{1,2,...,n}

a_kb_k

(here, we substituted k for π(k) in the sum, since π is a permutation of the set {1,2, ..., n}) and

n

X

k=1

|{z}

= P

k∈{1,2,...,n}

a_π(k)

n

X

i=k+1

| {z }

= P

i∈{1,2,...,n};

i>k

b_π(i)= X

k∈{1,2,...,n}

a_π(k) X

i∈{1,2,...,n};

i>k

b_π(i) = X

k∈{1,2,...,n}

X

i∈{1,2,...,n};

i>k

| {z }

= P

(i,k)∈{1,2,...,n}²; i>k

a_π(k)b_π(i)

= X

(i,k)∈{1,2,...,n}²; i>k

aπ(k)bπ(i)

). On the other hand, X

(i,k)∈{1,2,...,n}²; i>k

min

aπ(k)bπ(i), aπ(i)bπ(k)

= X

(k,i)∈{1,2,...,n}²; k>i

| {z }

= P

i∈{1,2,...,n}

P k∈{1,2,...,n};

k>i

= P

(i,k)∈{1,2,...,n}²; k>i

min

a_π(i)b_π(k), a_π(k)b_π(i)

| {z }

={^aπ(k)b_π(i),a_π(i)b_π(k)}

(here we renamed (i, k) as (k, i) in the sum)

= X

(i,k)∈{1,2,...,n}²; k>i

min

a_π(k)b_π(i), a_π(i)b_π(k) , (3)

and

2· X

(i,k)∈{1,2,...,n}²; i>k

min

a_π(k)b_π(i), a_π(i)b_π(k)

= X

(i,k)∈{1,2,...,n}²; i>k

min

aπ(k)bπ(i), aπ(i)bπ(k) + X

(i,k)∈{1,2,...,n}²; i>k

min

aπ(k)bπ(i), aπ(i)bπ(k)

= X

(i,k)∈{1,2,...,n}²; i>k

min

a_π(k)b_π(i), a_π(i)b_π(k) + X

(i,k)∈{1,2,...,n}²; k>i

min

a_π(k)b_π(i), a_π(i)b_π(k) (by (3))

= X

(i,k)∈{1,2,...,n}²; i6=k

min

a_π(k)b_π(i), a_π(i)b_π(k)

since the set

(i, k)∈ {1,2, ..., n}² | i6=k is the union of the two disjoint sets (i, k)∈ {1,2, ..., n}² | i > k and

(i, k)∈ {1,2, ..., n}² | k > i

= X

(i,k)∈{1,2,...,n}²; π(i)6=π(k)

| {z }

= P

i∈{1,2,...,n}

P k∈{1,2,...,n};

π(i)6=π(k)

min

a_π(k)b_π(i), a_π(i)b_π(k)

since i6=k is equivalent toπ(i)6=π(k) , because π is a permutation

= X

i∈{1,2,...,n}

X

k∈{1,2,...,n};

π(i)6=π(k)

min

a_π(k)b_π(i), a_π(i)b_π(k)

= X

i∈{1,2,...,n}

X

k∈{1,2,...,n};

i6=π(k)

min

a_π(k)bi, aib_π(k)

(here, we substitutedi for π(i) in the sum, since π is a permutation of the set {1,2, ..., n}))

= X

i∈{1,2,...,n}

X

k∈{1,2,...,n};

i6=k

| {z }

= P

(i,k)∈{1,2,...,n}²; i6=k

min{a_kb_i, a_ib_k}

(here, we substitutedk forπ(k) in the sum, sinceπ is a permutation of the set {1,2, ..., n}))

= X

(i,k)∈{1,2,...,n}²; i6=k

min{a_kb_i, a_ib_k}

Dividing this equation by 2, we obtain X

(i,k)∈{1,2,...,n}²; i>k

min

a_π(k)b_π(i), a_π(i)b_π(k) = 1 2

X

(i,k)∈{1,2,...,n}²; i6=k

min{a_kb_i, a_ib_k}. (4)

Now, for every pair (i, k)∈ {1,2, ..., n}² satisfying i > k, we have the inequality a_π(k)b_π(i) ≥min

a_π(k)b_π(i), a_π(i)b_π(k) , with equality if and only ifa_π(k)b_π(i) ≤a_π(i)b_π(k).

In other words,

a_π(k)b_π(i) ≥min

a_π(k)b_π(i), a_π(i)b_π(k) , (5)

with equality if and only if a_π(k)

b_π(k) ≤ a_π(i) b_π(i) (since a_π(k)b_π(i) ≤a_π(i)b_π(k) is equivalent to a_π(k)

b_π(k) ≤ a_π(i)

b_π(i)). Summing up the inequality (5) over all pairs (i, k)∈ {1,2, ..., n}² satisfying i > k, we obtain

X

(i,k)∈{1,2,...,n}²; i>k

a_π(k)b_π(i) ≥ X

(i,k)∈{1,2,...,n}²; i>k

min

a_π(k)b_π(i), a_π(i)b_π(k) ,

with equality if and only if

a_π(k)

b_π(k) ≤ a_π(i)

b_π(i) for every pair (i, k)∈ {1,2, ..., n}² satisfying i > k

.

In other words, X

(i,k)∈{1,2,...,n}²; i>k

a_π(k)b_π(i)≥ X

(i,k)∈{1,2,...,n}²; i>k

min

a_π(k)b_π(i), a_π(i)b_π(k) , (6)

with equality if and only if a_π(1)

b_π(1) ≤ a_π(2)

b_π(2) ≤...≤ a_π(n) b_π(n) (because

a_π(k)

b_π(k) ≤ a_π(i)

b_π(i) for every pair (i, k)∈ {1,2, ..., n}² satisfying i > k

is equiv- alent to a_π(1)

b_π(1) ≤ a_π(2)

b_π(2) ≤ ... ≤ a_π(n)

b_π(n)). Due to (4), we can rewrite the inequality (6) as

X

(i,k)∈{1,2,...,n}²; i>k

a_π(k)b_π(i)≥ 1 2

X

(i,k)∈{1,2,...,n}²; i6=k

min{a_kb_i, a_ib_k}, (7)

with equality if and only if a_π(1)

b_π(1) ≤ a_π(2)

b_π(2) ≤...≤ a_π(n) b_π(n) Upon addition of P

k∈{1,2,...,n}

akbk, the inequality (7) becomes

X

k∈{1,2,...,n}

a_kb_k+ X

(i,k)∈{1,2,...,n}²; i>k

a_π(k)b_π(i) ≥ X

k∈{1,2,...,n}

a_kb_k+1 2

X

(i,k)∈{1,2,...,n}²; i6=k

min{a_kb_i, a_ib_k},

(8) with equality if and only if a_π(1)

b_π(1) ≤ a_π(2)

b_π(2) ≤...≤ a_π(n) b_π(n). Due to (2), we can rewrite the inequality (8) as

n

X

k=1

a_π(k)

n

X

i=k

b_π(i)≥ X

k∈{1,2,...,n}

a_kb_k+1 2

X

(i,k)∈{1,2,...,n}²; i6=k

min{a_kb_i, a_ib_k},

with equality if and only if a_π(1)

b_π(1) ≤ a_π(2)

b_π(2) ≤...≤ a_π(n) b_π(n).

Thus, Theorem 1 is proven.

References

[1] J. G. Rau, W. O. J. Moser, R. J. Dickson, L. P. Prostanstus, Optimal Sequence of Products (problem E 2353 and solutions), American Mathematical Monthly vol. 80 (1973), pp. 437-438.