The wave function of a system of two identical particles For a system of two identical particles, where $r_1$ is the position vector of particle 1 and $r_2$ is the position vec. of particle 2, the wave function should be one of the plus or minus states: 
\begin{equation}
\psi _ \pm (r_1,r_2) = A [\psi _a(r_1) \psi_b (r_2) \pm \psi_b(r_1) \psi_a(r_2)] ,
\end{equation}
where $\psi_a$ and $\psi_b$ are the wave functions of particle 1 and 2 respectively [equation 5.10 of Griffith's Intro to Quantum Mechanics 2nd Ed.].
I see that this equation makes the wave function $\psi_\pm$ treat the two particles identically, but I don't know of any proof that it is actually the only way of writing this wave equation to treat them identically. For example, why not a wave function like:
\begin{equation}
\psi _ \pm (r_1,r_2) = A \sqrt{\psi^2 _a(r_1) \psi_b^2 (r_2) \pm \psi^2_b(r_1) \psi^2_a(r_2)}~?
\end{equation}
 A: The requirement is
$$
\psi(x_1,x_2) = 
\begin{cases}
  \psi(x_2,x_1) & \text{for bosons} \\
  -\psi(x_2,x_1) & \text{for fermions}.
\end{cases}
\tag{1}
$$
This property is required by the spin-statistics theorem in relativistic quantum field theory. Since non-relativistic quantum mechanics is supposed to be an approximation to relativistic quantum field theory, we also enforce it in non-relativistic QM.
A special case of equation (1) is
$$
\psi(x_1,x_2) \approx
\begin{cases}
  f(x_1)g(x_2)+f(x_2)g(x_1) & \text{for bosons} \\
  f(x_1)g(x_2)-f(x_2)g(x_1) & \text{for fermions},
\end{cases}
\tag{2}
$$
but like Lewis Miller's answer said, this is only a special case. The general requirement is equation (1).
The square-root example written in the question does not satisfy the requirement (1).
A: This product form of two-particle wave function is only correct if the particles are not interacting.  Nevertheless, it is often used as a first approximation, and if you take the expectation value of the true Hamiltonian and minimize it (by taking a variational derivative) you get the two-body Hartree-Fock equation which is often used to approximate the ground state energy and wave function for many-body systems of Fermions.  This approximation is often called the mean field approximation.
A: You’re forgetting that the wave function must also satisfy the TISE. With this condition the combine wavefunctions must be the sum of a permutation of products. 
A: If we have two particles, one in state $\psi_a$ and the other in state $\psi_b$, then the state vector would be $|\psi_a\rangle|\psi_b\rangle$. 
However, if the particles are indistinguishable, then it is equally likely to have the opposite be true (i.e. the "first" particle in state $\psi_b$ and the "second" particle in state $\psi_a$). Therefore, we would want the entire state to be a linear combination of these two states, each with equal weight. Therefore we end up with
$$|\Psi\rangle=|\psi_a\rangle|\psi_b\rangle\pm|\psi_b\rangle|\psi_a\rangle$$
If we choose to work in the $|r_1\rangle|r_2\rangle$ basis, then we end up with the expression you state.
I think the issue with your state is that it is not a "nice" linear combination of the states where one particle is in state $\psi_a$ and the other in state $\psi_b$. We need this if we want the postulate to hold that when $|\psi\rangle=\sum c_i|\psi_n\rangle$, we know that there is a probability of $|c_i|^2$ to measure the system to be in state $|\psi_i\rangle$
