k ≤ n, we obtain 1 4 X 1≤j<k≤n (aj+ ak

Problem 11670

(American Mathematical Monthly, Vol.119, November 2012) Proposed by Miranda Bakke, Benson Wu, and Bogdan Suceava (USA).

Prove that if n ≥ 3 and a1, . . . , an> 0, then (n − 1)

4

n

X

k=1

ak≥ X

1≤j<k≤n

ajak

aj+ ak

,

with equality if and only if all aj are equal.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

The inequality (a_j− a_k)²≥ 0 is equivalent to aj+ ak

4 ≥ ajak

aj+ ak

.

By summing over 1 ≤ j < k ≤ n, we obtain 1

4 X

1≤j<k≤n

(a_j+ a_k) ≥ X

1≤j<k≤n

a_ja_k aj+ ak

which yields the required inequality because 1

4 X

1≤j<k≤n

(aj+ ak) = (n − 1)

n

X

k=1

ak.

If all ajare equal then the equality holds. On the other hand, if aj₀ 6= ak₀ for some 1 ≤ j0< k0≤ n then (aj₀− ak₀)²> 0 and

a_j0+ a_k0

4 > a_j0a_k0 aj₀+ ak₀

.

This implies that in the required inequality the left-hand side is greater than right-hand side.