Ok, let me elaborate on my question. I really don't think the math is necessary as long as one understands all the physics behind the phenomenon, but I appreciate your skepticism.
1. A particle with mass ##m## (we can assume the electromagnetic mass is included here) and charge ##q## is moving with constant velocity ##v \ll c## along the x axis. The kinetic energy is ##\frac{1}{2} m v^2##. Since the EM mass is included in ##m##, the kinetic energy of the field is taken into account.
2. At ##t=0## we switch on a force ##F## acting opposite to the velocity and deccelerate the particle until its speed drops to ##0##. Of course, due to energy loss to radiation, the acceleration ##a \neq \frac{F}{m}##. Instead we have
$$ -F - \frac{q^2}{6 \pi \epsilon_0 c^3} \dot{a} = m a $$
Let ##t_0 = \frac{q^2}{6 \pi \epsilon_0 m c^3}##. The boundary conditions are ##a(0)=0##, ##v(0)=v##, ##v(T)=0## for some ##T## which depends on ##F##. The solution is
$$ a(t) = -\frac{F}{m} \left( 1- e^{-t/t_0} \right),$$
$$ F = \frac{mv}{T + t_0(e^{-T/t_0} -1 )}.$$
We use the Larmor formula to compute radiated energy:
$$ E_{rad} = \int_0^T m t_0 a(t)^2 dt.$$
For convenience, take ##v=m=t_0 = 1## - now ##F = \frac{1}{T -1 + e^{-T}}##. It is easy to see that ##F>0## and ##F \to \infty## as ## T \to 0##, which is intuitive. WolframAlpha says
$$ E_{rad} = \frac{T-\frac{3}{2} + 2e^{-T} - \frac{1}{2} e^{-2T}}{(T-1+e^{-T})^2}.$$
It goes to infinity as ##T \to 0##. In particular, ##\exists T>0 \ s.t. \ E_{rad} > \frac{1}{2} m v^2 ##. Moreover, there is an energy gathered from decceleration (negative work is done on the particle):
$$W = \int \vec{F} \cdot \vec{dx} < 0.$$
This is the maths. A simpler argument is this: radiated power is proportional to ##a^2##. This means, contrary to berkeman's claim, that integral of radiated power over time does depend on total time. This total time of decceleration can be arbitrarily short - although due to Abraham-Lorentz force the acceleration is not constant if we apply constant force. For very large force, acceleration will instead grow linearly, but there is no limit on the speed of this growth. Let ##a = -kt##. Then ##\int a(t)^2 dt = \frac{1}{3} k^2 T^3 = \frac{1}{3} k^2 \left( \frac{2v}{k} \right)^{3/2} \sim \sqrt{k}##.
