n λtr(A−1)+λtr(A−1) n n ≤ det A λ + λA−1

Problem 11907

(American Mathematical Monthly, Vol.123, April 2016) Proposed by X.-Q. Chang (USA).

LetA be an n × n positive-definite Hermitian matrix, with minimum and maximum eigenvalues λ andµ

 tr(A) µn + µn

tr(A)

ⁿ

≤ det A

µ + µA⁻¹

 ,

 n

λtr(A⁻¹)+λtr(A⁻¹) n

ⁿ

≤ det A

λ + λA⁻¹

 .

Solution proposed by Moubinool Omarjee, Lyc´ee Henri IV, Paris, France, and Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, Italy.

Any Hermitian matrix can be diagonalized by a unitary matrix: there exists a unitary matrix U such that A = U DU⁻¹where D is diagonal matrix with

λ = λ₁≤ λ2≤ · · · ≤ λⁿ= µ

on the main diagonal. Note that λ > 0 because A is positive-definite. Let t > 0 then

det A

t + tA⁻¹



= det

 U D

t + tD⁻¹

 U⁻¹



= det D

t + tD⁻¹



=

n

Y

k=1

 λk

t + t λ^k

 .

Let f (x) = ln ^x_t +_x^t
then by inspecting the sign of the second derivative

f^′′(x) = t⁴+ 4t²x²− x⁴ x²(x²+ t²)² , it is easy to see that f^tis convex in (0, tp

2 +√

5]. Therefore if µ ≤ tp 2 +√

5 then 1

nln

 det A

t + tA⁻¹



= 1 n

n

X

k=1

f (λ^k) ≥ f 1 n

n

X

k=1

λ^k

!

= f tr(A) n



= ln tr(A) tn + tn

tr(A)



which implies that

 tr(A) tn + tn

tr(A)

ⁿ

≤ det A

t + tA⁻¹

 .

Hence the first inequality follows by taking t = µ (notice thatp 2 +√

5 > 2).

Moroever the second inquality can be obtained from the first one by replacing A with A⁻¹ and µ

with 1/λ.