Problem 11512
(American Mathematical Monthly, Vol.117, June–July 2010) Proposed by F. Holland (Ireland).
LetN be a nonnegative integer. For x ≥ 0 prove that
N
X
m=0
1 m!
N −m+1
X
k=1
xk k
!m
≥1 + x + . . . + xN.
Solution proposed by Paolo Perfetti and Roberto Tauraso,
Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let P (x) denote the polynomial on the left-hand side of the inequality
P (x) = 1 +
N
X
m=1
1 m!
N −m+1
X
k1=1
. . .
N −m+1
X
km=1
xk1+...+km k1·k2· · ·km =
= 1 +
N
X
m=1
1 m!
m(N −m+1)
X
r=m
xr X
1≤k1,...,km k1+...+km=r
1 k1·k2· · ·km.
It is easy to verify that the degree of P (x) is N (N + 2)/4 ≥ N if N is even and (N + 1)2/4 ≥ N if N is odd. Hence, since the coefficients of P (x) are nonnegative, it suffices to show that the coefficients of the monomials of degree 0, 1, . . . , N of P (x), are equal to 1, that is
1 n!
dn dxnP (x)
x=0= 1 n = 0, 1, 2, . . . , N.
For n = 0 and for any value of N it is trivial, so we assume that 1 ≤ n ≤ N . Therefore
1 n!
dn
dxnP (x) = 1 n!
N
X
m=1
1 m!
m(N −m+1)
X
r=m
xr−n X
1≤k1,...,km k1+...+km=r
r(r − 1)(r − 2) · · · (r − n + 1) k1·k2· · ·km
and finally we have that 1
n!
dn dxnP (x)
x=0=
N
X
m=1
1 m!
X
1≤k1,...,km k1+...+km=n
1 k1·k2· · ·km
= [xn]
N
X
m=1
1 m!logm
1 1 − x
= [xn]
∞
X
m=1
1 m!logm
1 1 − x
= [xn]
exp
log
1
1 − x
−1
= [xn] x 1 − x = 1 .