For non-negative definite symmetric matrices, $\mathrm{tr}(AB)\le \mathrm{tr}(A)\mathrm{tr}(B)$

Question

$\DeclareMathOperator{\tr}{\mathrm{tr}}$ Is the inequality in title true for non-negative definite matrices?? I could neither prove this result, nor provide a counter example.

Context

I was trying to prove that Frobenius norm is a matrix norm, i.e. $$||AB||_F\le||A||_F\cdot||B||_F\\ \iff\tr(B^\top A^\top AB)\le\tr(A^\top A)\cdot\tr(B^\top B)\\ \iff\tr(A^\top ABB^\top)\le\tr(A^\top A)\cdot\tr(B B^\top)$$ Now if that happens for all positive definite matrices $A$ and $B$, that means trace of product of two non-negative definite matrices is less than or equal to the product of their traces.

Please help, and provide a proof of this result which doesn't rely on the multiplicative inequality for Frobenius norm (since that result is proved to be equivalent to it, and I don't know the proof of that result).

Answer 1

Let $(e_j)$ be an orthonormal basis of eigenvectors of $B$ and let $(\lambda_j)$ denote the corresponding eigenvalues. Then $$ \mathrm{tr}(AB)=\sum_j\langle AB e_j,e_j\rangle=\sum_j \lambda_j \langle A e_j, e_j\rangle\leq \max_j \lambda_j \mathrm{tr}(A)\leq \mathrm{tr}(A)\mathrm{tr}(B). $$