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.
Best Answer
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 $\le1$.