American Mathematical Monthly Problem 11434 by Slavko Simic, Belgrade, Serbia.
Let n be an integer such that n ≥ 2. Let x1, x2, ..., xn be distinct real numbers, and letp1, p2, ..., pn be positive numbers such that
n
P
k=1
pk= 1. Prove that min{x1, x2, ..., xn} ≤S ≤max{x1, x2, ..., xn},
where S denotes
n
P
k=1
pkx3k− n
P
k=1
pkxk 3
3
n
P
k=1
pkx2k− n
P
k=1
pkxk 2!.
Solution by Darij Grinberg.
First, let us show that min{x1, x2, ..., xn} ≤S.
In fact, let r = min{x1, x2, ..., xn}. Then, xi −r ≥ 0 for every i ∈ {1,2, ..., n}.
Hence,
n
X
k=1
pk(xk−r)3 ≥
n
X
k=1
pk(xk−r)
!3
(1) by the weighted Jensen inequality (applied to the positive reals x1 − r, x2 −r, ..., xn−r with the weights p1, p2, ..., pn), because the function x 7→x3 is convex on the nonnegative halfaxis. On the other hand,
n
X
k=1
pkx2k ≥
n
X
k=1
pkxk
!2
(2) by the weighted Jensen inequality (applied to the reals x1, x2, ..., xn with the weights p1, p2, ..., pn), because the function x7→ x2 is convex on the whole R. Now, a simple calculation (using only the definition of S and the binomial formula; we do not need to remember thatr = min{x1, x2, ..., xn}) shows that
S−r=
n
P
k=1
pk(xk−r)3− n
P
k=1
pk(xk−r) 3
3
n
P
k=1
pkx2k− n
P
k=1
pkxk
2! .
The right hand side of this equation is≥0 (since the numerator is≥0 by (1), and the denominator is ≥0 by (2)). Therefore, so is the left hand side. In other words, r≤S.
Since r= min{x1, x2, ..., xn}, this becomes
min{x1, x2, ..., xn} ≤S. (3)
Applying this to the reals −x1, −x2, ..., −xn instead of x1, x2, ..., xn, we get
min{−x1,−x2, ...,−xn} ≤S0, (4)
1
where
S0 =
n
P
k=1
pk(−xk)3− n
P
k=1
pk(−xk) 3
3
n
P
k=1
pk(−xk)2− n
P
k=1
pk(−xk) 2!
is the analogue ofS for the reals−x1,−x2, ..., −xn instead ofx1, x2, ..., xn. Obviously, S0 = −S and min{−x1,−x2, ...,−xn} = −max{x1, x2, ..., xn}, so that (4) becomes
−max{x1, x2, ..., xn} ≤ −S. In other words, S ≤ max{x1, x2, ..., xn}. Together with (3), this completes the solution of the problem.
2
