trace and spectral radius inequality to symmetric positive definite matrix

Question

If $\rho(A)$ is o spectral radius of $A$ and $tr(A)$ denotes the trace of $A$. The question is the following:

Let be $P,Q \in M_n(R)$ symmetric semidefinite positive matrix, then

$$tr(PQ)\le tr(P)\rho(Q)$$

I know that is necessary to use that the Frobenius and spectral norms are sub-multiplicative norms, and express the Frobenius norm how a sum of eigenvalues, but I have not idea yet. Help, please

Answer 1

Hint: $$ \operatorname{tr}PQ=\operatorname{tr}P^{1/2}P^{1/2}Q^{1/2}Q^{1/2}= \operatorname{tr}P^{1/2}Q^{1/2}Q^{1/2}P^{1/2}=\|Q^{1/2}P^{1/2}\|_F^2\le\ldots $$ Another way: $$ Q\le\rho(Q)\cdot I\implies P^{1/2}QP^{1/2}\le P^{1/2}\rho(Q)P^{1/2}=\rho(Q)\cdot P\implies\ldots $$