How to prove that if the expectation value of $A$ in any state is real, then $A$ is Hermitian? 
If the expectation value of operator $A$ in any state is real, then $A$ is Hermitian.

there is an uncompleted proof:
$$ \int(c_1\psi_1+c_2\psi_2)^* A (c_1\psi_1+c_2\psi_2)dx$$
$$=|c_1|^2\int\psi_1^* A \psi_1dx + |c_2|^2\int\psi_2^* A \psi_2dx
+c_1^*c_2\int\psi_1^* A \psi_2dx +c_2^*c_1\int\psi_2^* A \psi_1dx
$$
My teacher said that because 
$$c_1^*c_2\int\psi_1^* A \psi_2dx +c_2^*c_1\int\psi_2^* A \psi_1dx$$
is real, then we can conclude that
$$\int\psi_1^* A \psi_2dx = (\int\psi_2^* A \psi_1dx)^*$$
but I don't understand this last step, can someone explain it, or give another proof?
 A: Here is another version of the same proof.
If $\langle \psi | A \psi \rangle \in \mathbb R$ for all $\psi \in \cal H$, then
$\langle \psi | A \psi \rangle^* =  \langle \psi | A \psi \rangle$ for all $\psi \in \cal H$. Since $\langle \psi | \phi \rangle^* = \langle \phi| \psi\rangle$  we have that $\langle \psi | A \psi \rangle = \langle A\psi |  \psi \rangle$ that is $^1$ $\langle \psi | A \psi \rangle = \langle \psi | A^\dagger \psi \rangle$, namely:
$$\langle \psi | (A-A^\dagger) \psi \rangle =0 \quad \forall \psi \in {\cal H}\:.\tag{1}$$
Now consider $\psi = \phi + \chi$, where $\phi,\chi \in {\cal H}$ are arbitrary, obtaining:
$$\langle\phi + \chi| (A-A^\dagger) (\phi + \chi) \rangle =0 \tag{2}\:.$$
using the fact that $\langle \phi | (A-A^\dagger) \phi \rangle = \langle \chi | (A-A^\dagger) \chi \rangle 0$ in view of (1), and expanding (2) taking the real bilinerity of the scalar product into account we find
$$\langle \phi | (A-A^\dagger) \chi \rangle + \langle \chi | (A-A^\dagger) \phi \rangle=0\:.\tag{3}$$
The procedure can be implemented another time, starting again from (1) but now using $\psi = \phi + i\chi$. Exploiting the fact that the scalar product is anti linear in the left argument and linear in the right one, this time we end up with
$$i\langle \phi | (A-A^\dagger) \chi \rangle -i \langle \chi | (A-A^\dagger) \phi \rangle=0\:,$$
that is 
$$\langle \phi | (A-A^\dagger) \chi \rangle - \langle \chi | (A-A^\dagger) \phi \rangle=0\:,$$
Together with (3) it entails in particular that
$$\langle \chi | (A-A^\dagger) \phi \rangle =0 \quad \forall \phi, \chi \in {\cal H}\:.$$
We are free to choose $\chi =  (A-A^\dagger) \phi$ obtaining
$$||   (A-A^\dagger) \phi||^2 = \langle (A-A^\dagger) \phi | (A-A^\dagger) \phi \rangle =0 \quad \forall \phi \in {\cal H}\:,$$
which immediately implies 
$$ (A-A^\dagger) \phi =0 \quad \forall \phi \in {\cal H}\:.$$
Summing up, we have obtained
$$A^\dagger \phi = A\phi\quad  \forall \phi \in {\cal H}$$
that is the thesis. If the domain of $A$ is not the whole Hilbert space, the above reasoning holds however in that domain as it is a linear subspace by hypotheses.

(1) I just used the defintion of adjoint, $A^\dagger$, of $A$ that implies both
$\langle \psi | A \phi \rangle =  \langle A^\dagger \psi |  \phi \rangle$ and 
$\langle A\phi |  \psi \rangle =  \langle  \phi |  A^\dagger\psi \rangle$.
