Question: Referring to this diagram:

![enter image description here](https://i.stack.imgur.com/498wO.png)

I understand that any energy excess could be taken away as kinetic energy of the final $h$ and $Z$. If you are interested in this (kind of) process should you tune the collision energy at exactly $91+125=216~$GeV or is better to go up (a bit)?

I've found [this paper in arxiv](http://arxiv.org/abs/1411.0676) where they claim $\sim 250~$GeV. I would like to understand how to properly select the energy in case it matters.

Answer: 
In principle, you don't need to tune to 'exactly' this energy but having less than $m\_h + m\_Z$ suppresses this diagram. Having more should typically give you a higher cross section because there is 'more phase space' the final state particles can be in, i.e. the 'excess' energy will just be used as kinetic energy of the final state particles ($Z$ and $h$ 'move faster'). 

On the other hand, there is a competing effect of the $Z^\*$ in the s-channel being more off-shell (more away from the Z mass). The $Z$ propagator goes approximately like $\dfrac{1}{p\_Z^2 - M\_Z^2}$.

In the end, the cross section maximum is indeed somewhere around 250 GeV. Figure 7 of [arxiv:1308.6176](http://arxiv.org/abs/1308.6176) (reference 3 in the paper you refer to) shows this:

![Higgs cross sections in e+e- collisions](https://i.stack.imgur.com/d6h8P.png)
Completing the very good answer made by Andre Holzner, the electron and the positron in the initial state, have a non negligible probability to emit a photon with a significant energy. In particle physics jargon, this is called the ISR standing for Initial State Radiation. Therefore, you always have to be slightly above the threshold production to circumvent this problem.