Why does $2\to 2$ scattering cross-section have $E_{CM}^2$ in denominator? For $2\to 2$ scattering with equal masses we have
$$\left(\frac{d\sigma}{d\Omega}\right)_{CM} = \frac{1}{64 \pi^2 E_{CM}^2} |\mathcal{M}|^2.$$
(Schwartz's QFT eq. 5.33) Can we make the general statement that total cross section increases as $E_{CM}$ decreases for scattering processes? I'm guessing not because $\mathcal{M}$ may have a complicated dependence on energy/momenta.

As an example, take the case of $e^+e^-\to \mu^+\mu^-$. With $m_e = 0$ we have
$$\frac{d\sigma}{d\Omega} = \frac{\alpha^2}{4 E_{CM}^2} \sqrt{1-m_{\mu}^2/E_{\mu}^2} \left(1+ m_{\mu}^2/E_{\mu}^2 + (1- m_{\mu}^2/E_{\mu}^2)\cos^2\theta\right)$$
in the c.o.m frame (Schwartz 13.78). Here it's pretty clear that total cross section increases as energy decreases.

But I would think that particles scatter more when they are given more energy? The limiting case of course being zero velocity $\to $ zero scattering. I would appreciate any intuition on the relation between $E_{CM}$ and total cross section.
 A: 
Can we make the general statement that total cross section increases
  as $E_{CM}$ decreases for scattering processes?

If the considered QFT is unitary (at least perturbatively), there is a constraint on the cross-section following from the unitarity, called Froissart bound. It is the direct consequence of the optical theorem. The latter states that
$$
\sigma_{1+2\to\text{all}} \simeq \frac{1}{s}\text{Im}(M_{\text{forward}}),
$$
where $\sigma_{1+2\to\text{all}}$ is the $1+2\to \text{all}$ scattering cross-section, while $s$ is the invariant mass of the colliding particles. The amplitude $M_{\text{forward}}$ is the forward scattering amplitude bounded by
$$
M_{\text{forward}}\lesssim s\ln^{2}(s)
$$
Therefore, for general $1+2\to\text{all}$ scattering one obtains the Frossairt bound:
$$
\sigma_{1+2\to \text{all}} \lesssim \ln^{2}(s)
$$

But I would think that particles scatter more when they are given more
  energy?

The more energy of the particle is, the less corresponding de Broglie wave-length is. Therefore qualitatively it's rather expected that the colliding particles with very large energies will overlook each other.
