Eigenvalues of positive definite matrices Let $A, B$ be positive definite matrices. What can be said about the maximum eigenvalue of $B^{1/2}AB^{1/2}$ in terms of the eigenvalues of $A$ and $B$?
 A: What can be said: as Omnomnomnom mentioned, it is less than or equal the product of the largest eigenvalues of $A$ and $B$. Other than that, not a lot can be said. 
For instance, you could have 
$$
A=\begin{bmatrix}2&0\\0&1\end{bmatrix},\ \ \ B=\begin{bmatrix}1&0\\0&3\end{bmatrix}.
$$
Here $\lambda_1(A)=2$, $\lambda_2(B)=3$, $\lambda_1(AB)=3$.
But if
$$
A=\begin{bmatrix}2&0\\0&1\end{bmatrix},\ \ \ B=\begin{bmatrix}3&0\\0&1\end{bmatrix},
$$
now $\lambda_1(A)=2$, $\lambda_2(B)=3$, $\lambda_1(AB)=6$.
Or if 
$$
A=\begin{bmatrix}2&0\\0&1/10\end{bmatrix},\ \ \ B=\begin{bmatrix}1/10&0\\0&3\end{bmatrix},
$$
now $\lambda_1(A)=2$, $\lambda_2(B)=3$, $\lambda_1(AB)=0.3$. Pushing this idea one can make $\lambda_1(AB)$ as close to $0$ as desired, without changing $\lambda_1(A)$ nor $\lambda_1(B)$. 
A: Let $\|A\|$ denote the spectral norm of $A$ (which gives the largest eigenvalue for positive definite matrices).  We have
$$
\|B^{1/2}AB^{1/2}\| \leq
\|B^{1/2}\| \cdot \|A\| \cdot \|B^{1/2}\| = \|A\|\|B\|
$$
that is, it is at most the product of the largest eigenvalues of $A$ and $B$.
