Symmetric matrices with the eigenvalues comparable Let $A,B$ be $n\times n$ real symmetric matrices, with eigenvalues $\lambda_i$ and $\mu_i$ respectively, $i=1,\cdots,n$. Suppose that 
$$\lambda_i\leq\mu_i,\forall\ i.$$
Show that there exists an orthogonal matrix $O$ such that 
$$O^TBO-A$$
is non-negative definite.
I do want to show that for some orthogonal matrix $P$ such that $P^tAP$ commutes with $B$, but this idea could not be forwarded...

Let $P,Q$ be orthogonal with $P^tAP$ the diagonal matrix with diagonal $\lambda_1,\dots,\lambda_n$ and $Q^tBQ$ the diagonal matrix with diagonal $\mu_1,\dots,\mu_n$. Then $Q^tBQ-P^tAP$ is non-negative definite, so $PQ^tBQP^t-A$ is. Let $O=QP^t$. 
