Problem 11759
(American Mathematical Monthly, Vol.121, February 2014) Proposed by O. Ganea (Switzerland) and C. Lupu (USA).
LetA be an n × n skew-symmetric real matrix.
Show that for positive real numbersx1, . . . , xk withk ≥ 2,
det(A + x1I) · · · det(A + xkI) ≥ (det(A + (x1· · ·xk)1/kI))k. In addition, show that if also allxi lie on the same side of1, then
det(A + I)k−1det(A + (x1· · ·xk)I) ≥ det(A + x1I) · · · det(A + xkI).
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Since A is a n × n skew-symmetric real matrix, it follows that there exists a n × n orthogonal matrix Q, and real numbers λ1, . . . λr, such that
A = Q · diag(B(λ1), . . . , B(λr), 0, · · · , 0) · Q−1 where B(λi) =
0 λj
−λj 0
,
which implies that det(A + tI) = tn−rQr
j=1(t2+ λ2j). Thus, the first inequality becomes
(x1· · ·xk)n−r
k
Y
i=1 r
Y
j=1
(x2i + λ2j) ≥
(x1· · ·xk)(n−r)/k
r
Y
j=1
((x1· · ·xk)2/k+ λ2j)
k
,
that is
r
Y
j=1 k
Y
i=1
(x2i + λ2j) − ((x21· · ·x2k)1/k+ λ2j)k
!
≥0,
which holds because fj(t) = ln(et+ λ2j) is convex and
ln
k
Y
i=1
(x2i + λ2j)
!
=
k
X
i=1
fj(ln(x2i)) ≥ kfj(
k
X
i=1
ln(x2i)/k) = ln
((x21· · ·x2k)1/k+ λ2j)k .
In a similar way, the second inequality becomes
r
Y
j=1
(1 + λ2j)
k−1
(x1· · ·xk)n−r
r
Y
j=1
((x1· · ·xk)2+ λ2j)
≥(x1· · ·xk)n−r
k
Y
i=1 r
Y
j=1
(x2i + λ2j)
that is
r
Y
j=1
x21· · ·x2k+ λ2j 1 + λ2j −
k
Y
i=1
x2i + λ2j 1 + λ2j
!
≥0
which holds by the following argument. The function gj(t) = fj(t) − ln(1 + λ2j) is convex, increasing and gj(0) = 0 which implies that it is superadditive in [0, +∞) and in (−∞, 0].
Hence, if all xi lie on the same side of 1 then all ln(x2i) lie on the same side of 0, and
ln x21· · ·x2k+ λ2j 1 + λ2j
!
= gj k
X
i=1
ln(x2i)
!
≥
k
X
i=1
gj(ln(x2i)) = ln
k
Y
i=1
x2i + λ2j 1 + λ2j
! .
