Problem 11826
(American Mathematical Monthly, Vol.122, March 2015)
Proposed by M. Bataille (France).
Let m and n be positive integers with m ≤ n. Prove that
n
X
k=m
4n+1−km + k − 1 m − 1
2
≥
n
X
k=m
m + n k
2
.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
For 1 ≤ m ≤ n, let
Fm(n) =
n
X
k=m
1 4k
m + k − 1 m − 1
2
− 1 4n+1
n
X
k=m
m + n k
2 .
Then
Fm(m) = 1 4m
2m − 1 m − 1
2
− 1 4m+1
2m m
2
= 0.
So it suffices to show that Fm(n) ≥ Fm(n − 1) for n > m. Note that
n
X
k=m
m + n k
2 k m + n =
n
X
k=m
m + n m + n − k
2m + n − k m + n =
n
X
k=m
m + n k
2
−
n
X
k=m
m + n k
2 k m + n
implies that
n
X
k=m
m + n k
2
2k m + n =
n
X
k=m
m + n k
2
.
Hence
Fm(n) − Fm(n − 1) = 1 4n
m + n − 1 m − 1
2
− 1 4n+1
n
X
k=m
m + n k
2 + 1
4n
n−1
X
k=m
m + n − 1 k
2
= 1 4n
n−1
X
k=m−1
m + n − 1 k
2
− 1 4n+1
n
X
k=m
m + n k
2
= 1 4n+1
n
X
k=m
m + n k
2 2k m + n
2
−
n
X
k=m
m + n k
2!
≥ 0,
because by Cauchy-Schwarz inequality
n
X
k=m
m + n k
2
·
n
X
k=m
m + n k
2 2k m + n
2
≥
n
X
k=m
m + n k
2 2k m + n
!2
=
n
X
k=m
m + n k
2!2
.
