Multiplicative property of trace Let $T_1 \in \mathcal{L}(V)$ and $T_2 \in \mathcal{L}(V)$ be positive operators. Prove that the trace of their product is non-negative i.e., tr($T_1 T_2) \geq 0$
Attempt 1: Obviously, a positive operator has a positive trace because all of its eigenvalues are positive. However, the product of two positive operators is not necessarily positive.
Attempt 2: Finding a pattern between the positivity of eigenvalues during multiplication
Both of my attempts fizzled out and I am pretty stuck. How would one solve this?

The product of two positive matrices $A,B \ge 0$ always has nonnegative eigenvalues. Indeed, since $B$ is positive it has a unique positive square root $B^{1/2}$ so
$$\sigma(AB) = \sigma(AB^{1/2}B^{1/2}) = \sigma(B^{1/2}AB^{1/2}).$$
Notice that $B^{1/2}AB^{1/2}$ is a positive matrix since it is self-adjoint and for all $x \in V$ holds
$$\langle B^{1/2}AB^{1/2}x, x\rangle = \langle AB^{1/2}x, B^{1/2}x\rangle \ge 0$$
because $A \ge 0$. In particular, $B^{1/2}AB^{1/2}$ has nonnegative spectrum and hence the same holds for $AB$.
Since the trace is the sum of eigenvalues, it follows that $\operatorname{Tr}(AB) \ge 0$.
