A sort of AM-GM inequality for matrices

Question

Let $A, B$ be symmetric, positive definite matrices. Is it true that $$ tr \left( (A B^2 A)^{1/2}\right) \leq \frac{1}{4} \| A + B\|_F^2? $$ In the diagonal case, the left-hand side is $tr(AB)$ and, denoting by $\lambda_i$ and $\mu_i$ the diagonal elements of $A$ and $B$, the inequality follows from the usual AM-GM inequality: $$ tr(AB) =\sum_i \mu_i \lambda_i \leq \frac{1}{4} \sum_i (\mu_i + \lambda_i)^2 = \frac 14 \| A + B\|_F^2. $$

To convince ourselves that the inequality is true, we can run the following python code, which prints 0.99999375.

import numpy as np
import scipy.linalg as la

n, d, ratios = 100000, 3, []
for i in range(n):
    A = np.random.randn(d, d)
    B = np.random.randn(d, d)
    A, B = A.dot(A.T), B.dot(B.T)
    lhs = np.matrix.trace(la.sqrtm(A.dot(B).dot(B).dot(A)))
    rhs = (1/4)*la.norm(A + B, ord='fro')**2
    ratios.append(lhs/rhs)

print(np.max(ratios))

After a little literature search, it seems that a slightly modified version of the inequality, with a factor $1/2$ instead of $1/4$ in the right-hand side, is a consequence of Theorem IX.4.2 in the textbook Matrix Analysis by Rajendra Bhatia.

Statement of Theorem IX.4.2: for any two matrices, $$s_j(A^*B) \leq \frac{1}{2} \, s_j(AA^* + BB^*),$$ where $s_j$, $j = 1, \dots, n$, denotes the $j$-th singular value.

Proof that this partially answers the question: we can rewrite the left-hand side as $$ GM(A, B) := tr \left( (A B^2 A)^{1/2}\right) = tr \left(\left((BA)^*BA\right)^{1/2} \right) = \sum_j s_j(BA). $$ Using the quoted result, we therefore obtain \begin{align} GM(A, B) &\leq \frac{1}{2} \sum_j s_j(A^2 + B^2) = \frac{1}{2} tr(A^2 + B^2) \\ &= \frac 14 tr((A + B)^2 + (A - B)^2) \\ &= \frac 14 tr(A + B)^2 + \frac{1}{4} tr(A - B)^2 \\ &= \frac 14 \|A + B\|_F^2 + \frac 14 \|A - B\|_F^2 \\ &\leq \frac 12 \|A + B\|_F^2 =: 2 \, AM(A, B). \end{align} Let us note that numerical experiments show that the inequality $$ \frac{1}{2} \sum_j s_j(A^2 + B^2) \leq 2 \, AM(A, B), $$ is sharp.

Answer 1

It turns out that this question is non-trivial and was resolved only in 1999, by Rajendra Bhatia and Fuad Kittaneh in the paper Notes on matrix arithmetic–geometric mean inequalities, freely available here. They show that, for any unitarily invariant matrix norm $\|.\|$, $$ \| A B\|\leq \frac 14 \|(A+B)^2\|. $$ In the particular case of the Schatten norm with $p = 1$, this means $$ \sum_j s_j(AB) \leq \frac{1}{4} \sum_j s_j ((A + B)^2) = \frac{1}{4}\sum_j (s_j(A+B))^2, $$ which is the inequality we were trying to show.

Answer 2

Denote $C = AB$, then $CC^t$ is positive definite. Hence,

$$ tr \left( (A B^2 A)^{1/2} \right) = tr \left( (CC^t)^{1/2} \right) \leq \left( tr(CC^t) \right)^{1/2} = \left( tr (A B^2 A)\right)^{1/2} = \left( tr (A^2 B^2 )\right)^{1/2} = tr(AB).$$

Now,

$$ tr(AB) = \sum_i \sum_j a_{i,j} b_{j,i} \leq \sum_i \sum_j \frac{1}{4}(a_{i,j} + b_{j,i})^2 = \sum_i \sum_j \frac{1}{4}(a_{i,j} + b_{i,j})^2 = \frac{1}{4} \| A + B\|_F^2.$$

Hence the result