Problem 11769
(American Mathematical Monthly, Vol.121, April 2014) Proposed by P. P. D´alyay (Hungary).
Leta1, . . . , an and b1, . . . , bn be positive real numbers. Show that
n
X
j=1
aj
bj
2
− 2
n
X
j,k=1
ajak
(bj+ bk)2 ≤ 2√ 2
n
X
j,k=1
ajak
(bj+ bk)
n
X
j,k=1
ajak
(bj+ bk)3
1/2
.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
For x ∈ R, let f(x) =Pn
j=1aje−bjx. Then, for any non-negative integer m, C :=
Z ∞
0
f (x) dx =
n
X
j=1
aj
bj, Dm:= 1 m!
Z ∞
0
xmf2(x) dx =
n
X
j,k=1
ajak
(bj+ bk)m+1, By Cauchy-Schwarz inequality, for t > 0,
Z ∞ 0
f (x)dx
2
=
Z ∞ 0
1
t + x(t + x)f (x)dx
2
≤
Z ∞ 0
dx (t + x)2
Z ∞ 0
(t + x)2f2(x)dx
= t Z ∞
0
f2(x)dx + 2 Z ∞
0
xf2(x)dx + 1 t
Z ∞
0
x2f2(x)dx.
that is
C2− 2D1≤ tD0+2D2
t . By letting t =p2D2/D0> 0, we obtain C2−2D1≤ 2√
2√
D2D0which is equivalent to the required
inequality.
The original statement of the problem is false. Our correction is based on the paper A Reverse Hilbert-like Optimal Inequalityby Omran Kouba (arXiv:1404.6744).
