$\langle u| \langle v| A |u\rangle |v\rangle \geq 0$ for any $|u\rangle \in \mathcal{H}_1$ and $|v\rangle \in \mathcal{H}_2$. Then $A$ is Hermitian? $\mathcal{H}_1$ and $\mathcal{H}_2$ are complex Hilbert spaces. $A$ is an operator from $\mathcal{H}_1\otimes \mathcal{H}_2$ to $\mathcal{H}_1\otimes \mathcal{H}_2$. 
Suppose that $\langle u| \langle v| A |u\rangle |v\rangle \geq 0$ for any vectors $|u\rangle \in \mathcal{H}_1$ and
$|v\rangle \in \mathcal{H}_2$. Then $A$ is Hermitian. 
The source of the above statement is M . Horodecki, P. Horodecki, and R. Horodecki, “On the necessary and sufficient conditions for separability of mixed quantum states” (e-print archive: https://arxiv.org/abs/quant-ph/9605038) 
Attempt of Proof.
$\langle u|\langle v| A |u\rangle |v\rangle  \in \mathbb{R}$.
$\langle u|\langle v| A |u\rangle |v\rangle =(\langle u|\langle v| A |u\rangle |v\rangle )^* = \langle u|\langle v| A^{\dagger} |u\rangle |v\rangle  $
$\langle u|\langle v| A-A^{\dagger} |u\rangle |v\rangle  = 0 $ for any $|u\rangle \in \mathcal{H}_1$ and
$|v\rangle \in \mathcal{H}_2$.
Now I am lost how to proceed to show that $A = A^{\dagger} $. 
Any hints or references are appreciated. Thank you. 
 A: Claim: For an arbitrary Hermitian operator $M$, there exist projections $P_j,Q_j$ and real coefficients $a_j$ such that
$$
M = \sum_{j}a_j P_j \otimes Q_j.
$$
Proof of Claim: Note that $M$ can be written in the form $M = \sum_p A_p \otimes B_p$, where $A_p,B_p$ are Hermitian matrices (this can be proven formally, but it is easy to see in terms of matrices if we take $\{A_p\}$ to be a nice (sparse) basis of the Hermitian matrices).  Moreover, we can write $A_p = \sum_q a_{pq} P_q$ for real coefficients $a_{pq}$ (for instance, using a spectral decomposition) and similarly $B_p = \sum_{q}b_{pq}Q_q$.  With that have
$$
\begin{align}
M &= \sum_p A_p \otimes B_p = \sum_{pq} a_{pq}b_{pq}\,A_{pq} \otimes B_{pq}.
\quad \square
\end{align}
$$
Now, we can prove the result as follows: suppose that $M$ is an operator satisfying $\operatorname{Tr}(M(P \otimes Q)) \geq 0$ for all projections $P,Q$.  To show that $M$ is Hermitian, it suffices to show that for any $w \in \mathcal H_1 \otimes \mathcal H_2$, we have $w^*Mw \in \Bbb R$.  Because $ww^* \in L(\mathcal H_1 \otimes \mathcal H_2)$ is a Hermitian operator, there exist real coefficients $a_j$ such that 
$$
ww^* = \sum_{j} a_j P_j \otimes Q_j.
$$
It follows that
$$
w^*Mw = \operatorname{Tr}(M ww^*) = 
\operatorname{Tr}\left(M \sum_{j}a_j P_j \otimes Q_j\right)
= \sum_j a_j \operatorname{Tr}(M(P_j \otimes Q_j)) \in \Bbb R
$$
as desired.
