Is there any atom which is dia-electric? Take an atom. Suppose we impose some magnetic field on it. For some atoms, the energy increases---this is a phenomenon of diamagnetism. 
The question is, how about an electric field? Can the energy of the atom increase when the electric field is turned on? 
Put in a different way, can the electric field induce a dipole anti-parallel to it? If not, why can a magnetic field? 
 A: No, there can't be atoms like that, at least not in the real world.
In the magnetic case, diamagnetism means that the magnetic susceptibility may be negative (so the permeability may be lower or higher than in the vacuum, the magnetic susceptibility may have both signs).
But in the electric case, the electric susceptibility is always positive, and the permittivity of a material is always greater than the permittivity of the vacuum.
The asymmetry arises because atoms are full of charged particles that move in the expected direction. On the other hand, they don't contain any magnetic monopoles, just dipoles, and their behavior is harder to predict, if I omit the detailed explanations of paramagnetism, diamagnetism etc.
A: Possibly it can be proven in this way. Let $H_0$ denote the unperturbed Hamiltonian, and let $D_z = \sum_{i=1}^N z_i $ denote the $z$-component of the electric dipole. 
Define 
$$ H (f) = H_0 - f D_z . $$
Let $E_g(f)$ be the ground state energy of $H(f)$. First, apparently, $E_g$ is an even function of $f $.
Let $|\psi(f)\rangle $ be the ground state of $H(f)$. We have 
$$ E_g(0) = \langle \psi(0)|H(0)|\psi(0)\rangle =\frac{1}{2} \langle \psi(0)|H(f)|\psi(0)\rangle + \frac{1}{2} \langle \psi(0)|H(-f)|\psi(0)\rangle  \\
 \geq  \frac{1}{2} \langle \psi(f)|H(f)|\psi(f)\rangle + \frac{1}{2} \langle \psi(-f)|H(-f)|\psi(-f)\rangle \\
= \frac{1}{2} E_g(f) + \frac{1}{2} E_g (-f) = E_g(f) . \\$$
QED. 
Anyway, the point is that, the ground state energy as a function of $f$ is a concave function. 
