Problem 12066
(American Mathematical Monthly, Vol.125, October 2018) Proposed by Xiang-Qian Chang (USA).
Let n and k be integers greater than 1, and let A be an n × n positive definite Hermitian matrix.
Prove
(det(A))1/n≤ tracek(A) − trace(Ak) nk− n
1/k .
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. The matrix A is diagonalizable and all its eigenvalues λ1, λ2, . . . , λn are real and positive numbers. Hence the given inequality is equivalent to
n
Y
j=1
λkj
1/n
≤
Pn j=1λjk
− Pn
j=1λkj
nk− n . (⋆)
Now the right-hand side of (⋆) is an arithmetic mean:
Pn j=1λj
k
− Pn
j=1λkj
nk− n = 1
|S|
X
j∈S
λj1λj2· · · λjk
where S is the set of k-ple j = (j1, j2, . . . , jk) ∈ {1, 2, . . . , n}k such that ja6= jb for some a 6= b.
By the AM-GM inequality, this arithmetic mean is greater than or equal to the corresponding geometric mean. Since each index j appears as the i-th component of j ∈ S exactly nk−1− 1 times, we have that such geometric mean can be written as
Y
j∈S
λj1λj2· · · λjk
1/|S|
=
n
Y
j=1
(λnjk−1−1)k
1/(nk−n)
=
n
Y
j=1
λkj
1/n
which is just the left-hand side of (⋆).
