Why bound states in QFT have higher mass than single particle states? 
In standard textbooks in QFT while discussing e.g. the Kallen-Lehmann formula (see e.g. Section 7.1 in the Peskin-Schroeder book) it is always assumed that bound states of two or more particles have higher mass than the one particle states.
  Why this should be true?

Let us compare this for example with the classical two particles interacting according to the Coulomb law.They may rotate around each other  with fixed distance $R$ such that the center of mass is at rest, thus forming a bound state. The total energy goes to $-\infty$ when $R$ goes to 0.
Analogous problem arises in the case of two non-relativistic quantum particles interacting again according to Coulomb law.  While the total energy of the bound state cannot be arbitrarily small as in the classical case, the discrete energy levels (corresponding to bound states) are negative. At the same time if the particles are at rest and are far apart from each other and thus do not interact, their energy vanishes. 
Am I wrong?
 A: The two(or more)-particle states you are referring to are not bound states. They are scattering states of multiple particles. Recall that irreducible representations $[m,s]$ of the Lorentz group are labeled by two Casimir operators, the invariant mass:
$$ M^2 = - P_\mu P^\mu \ ,$$
where $P_\mu$ is the energy-momentum vector and the eigenvalues will be denoted by $m^2$; and
$$ - W^\mu W_\mu \ ,$$
where $W_\mu$ is the Pauli-Lubanski pseudovector, but this is just for completeness (this has eigenvalues $s(s+1)$) .
Now the one-particle states that the QFT books consider are simply the representation with just one $[m,s]$. The two-particle states are states $[m,s] \otimes [m',s']$ etc. If we have such a many-particle representation, we can decompose it into irreducibles, but we will get a continuous family of representations from $M = m+m'$ to $\infty$.
Bound states, on the other hand, are one-particle representations with some definite $m$. That such a bound states may be formed from other particles simply means that there is a transition matrix element.
In particular, to get an unitary S-matrix, one has to assume that all bound states are included as particle states in the asymptotic Hilbert spaces.
