Problem 11680
(American Mathematical Monthly, Vol.119, December 2012) Proposed by Benjamin Bogosel (France), and Cezar Lupu (Romania).
Let x1, . . . , xn be nonnegative real numbers. Show that
n
X
i=0
xi i
!4
≤ 2π2
n
X
i,j=1
xixj i + j
n
X
i,j=1
xixj (i + j)3.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
It suffices to apply the following integral version of Carlson’s inequality: if f is a non negative function defined in [0, +∞) such that f (t), tf (t) ∈ L2([0, +∞) then
Z +∞
0
f (t) dt
4
≤ π2 Z +∞
0
(f (t))2dt Z +∞
0
t2(f (t))2dt.
Indeed, take f (t) =Pn
i=1xie−it, then, since for any nonnegative a Z +∞
0
e−atdt = 1 a and
Z +∞
0
t2e−atdt = 2 a3, it follows that
Z +∞
0
f (t) dt =
n
X
i=1
xi
Z +∞
0
e−itdt =
n
X
i=0
xi
i , Z +∞
0
(f (t))2dt =
n
X
i,j=1
xixj
Z +∞
0
e−(i+j)tdt =
n
X
i=0
xixj
i + j, Z +∞
0
t2(f (t))2dt =
n
X
i,j=1
xixj
Z +∞
0
t2e−(i+j)tdt = 2
n
X
i=0
xixj (i + j)3,
and the required inequality holds by the above Carlson’s inequality.
