Proof of a matrix trace inequality

Question

Suppose $A$ and $B$ are both positive semidefinite matrices with the same size. Prove the following inequality: \begin{align} \text{Tr}((A^{1/2} B A^{1/2})^{1/2}) \geq \text{Tr}(A^{1/2}B^{1/2}), \end{align} where the square root of a PSD matrix is defined to be the principal square root, which is also a PSD matrix.

Clearly, equality holds when $A$ and $B$ commute.

Answer 1

This follows from the inequality that $\operatorname{tr}\left((XX^\ast)^{1/2}\right)\ge|\operatorname{tr}(X)|$. Let $X=PU$ be the polar decomposition of $X$, where $P$ is positive semidefinite and $U$ is unitary. The inequality then reduces to $\operatorname{tr}(P)\ge|\operatorname{tr}(PU)|$, which is obviously true because all diagonal entries of a unitary matrix have moduli $\le1$.

Answer 2

Hint: if $X$ is psd, then $(\mathrm {Tr}\, X)^2 \geq \mathrm{Tr}\, X^2$. This can be proven by elementary inequalities if you consider the above as functions of the eigenvalues of $X$.

Your inequality follows from appropriate choice of $X$.