Entanglement of Tripartite States Is there any simple algorithm to determine the entanglement of a tri-partite state?
In particular, what is the proof for entanglement of $ |GHZ\rangle $ and $ |W\rangle $ states?
$ |GHZ\rangle =\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle) $
$ |W\rangle =\frac{1}{\sqrt{3}}(|001\rangle+|010\rangle+|100\rangle) $ 
 A: You can calculate the density matrix of the state $$\rho = |\varphi\rangle\langle\varphi|$$ And then, the reduced density matrix for one of the particles taking the partial trace $$ \rho_A = \textrm{tr}_{B,C}(\rho)$$ The state is entangled if and only if the reduced matrix is a mixed state. This can be checked, for example, calculating the von Neumann entropy $$ S  = -\textrm{tr} (\rho_A \log \rho_A)$$
A: $\newcommand{\ket}[1]{\left| #1 \right>}$A state  $ \ket \psi \in H_1 \otimes H_2 \otimes H_3$ is said to be entangled if there exist no coefficients $a_i,b_i,c_i$ such that:
$$\ket \psi = \sum_{ijk} d_{ijk} \ket{e^1_i} \otimes \ket{e^2_j} \otimes \ket{e^3_j} = \sum_i a_i \ket{e^1_i} \otimes \sum_j b_j \ket{e^2_j}   \sum_k c_k \ket{e^3_k} \tag{1}$$
where $\ket{e^i_j}$ is the $j$th basis vector of $H_i$.
If there exists such coefficients than the state said to be separable and the state is not entangled. 
People tend to condense the notation by defining $\ket{e_1e_2e_3}\equiv \ket{e_1} \otimes \ket{e_2} \otimes \ket{e_3}$ so I shall do so in what is next to come. 
Assuming that your states are doublet states, let's start calculating. I'll only do the first one, which is easier to do and leave the second one to you, which is analogous to what I'll do now.
Assume that you can write your state in the separable  form (1)
\begin{align}
\ket{GHZ}&= \frac{1}{\sqrt{2}} (\ket{000}+\ket{111})\\&\overset{?}{=} \big(a_1 \ket 0 + a_2 \ket 1 \big) \otimes \big(b_1 \ket 0 + b_2 \ket 1 \big) \otimes \big(c_1 \ket 0 + c_2 \ket 1 \big) \\
&= a_1b_1c_1\ket{000}+ a_2b_2c_2 \ket{111} + \text{other linear combinations}
\end{align} 
Notice that in order to get your state you cannot have any of the $a_i,b_i$ or $c_i$ to be zero, if this would be the case you cannot get either $\ket{000}$ or $\ket{111}$. However these are all the coefficients you have, therefore if none of them zero, then you cannot get any of the other linear combinations to vanish, which is a contradiction to the assumption that you can write your state in the separable form (1). Thus this state is an entangled state.
Note that the intuition that I talked about in my comment above saves you a lot of time and effort but as  Ellis Cooper said, "Rigor cleans the window through which intuition shines.", so it is up to you how much rigour you want.
