Let $\rho \in \mathfrak{L}(A)$ and $\sigma \in \mathfrak{L}(A)$. Now I want to know can we say that
for all $\sigma \geqslant 0:$$\quad \quad Tr(\rho \sigma) \geqslant 0 \quad \quad$ if and only if $\quad \quad\rho \geqslant 0 $
$\rho \geqslant 0 $ and $\sigma \geqslant 0 $ means that both of them are positive semi-definite operator and $Tr$ means trace. $A$ is a Hilbert space and $\mathfrak{L}(A)$ means all linear maps on $A$.
Could you please help me how can I show it?
The "if" direction does hold:
Let $\rho,\sigma$ be positive-semidefinite (matrices, linear operators). Denote by $\rho^{1/2}$ the positive-semidefinite square root of $\rho$ ("the", as it is uniquely determined). Then $$\operatorname{Tr}(\rho\sigma) \;=\; \operatorname{Tr}(\rho^{1/2}\sigma\rho^{1/2})\;\geqslant\; 0$$ by cyclicity of the trace, and because $\,\rho^{1/2}\sigma\rho^{1/2}\,$ is positive-semidefinite.
Regarding the other direction, the "only if": $$\text{With}\quad\sigma = \begin{pmatrix}1&0\\0&1\end{pmatrix},\quad \rho =\begin{pmatrix}2&0\\0&-1\end{pmatrix} \text{ you get } \operatorname{Tr}(\rho\sigma)=1\,.$$