How to calculate the bremsstrahlung limit in the fusion triple product diagram? The fusion triple product is a commonly used figure of merit to quantify the progress in fusion research, it is the product of plasma density $n$,  temperature $T$ and energy confinement time $\tau_E$. 
It can be used in combination with a power balance to formulate a criterion for ignition, which is defined as the point when the energy released by the fusion reactions is large enough to sustain them (by proving enough heat to keep the fusion processes going). The criterion reads
$$
nT\tau_e > \frac{12\,T^2}{\left<\sigma_{fusion}u\right>\epsilon_\alpha - 4c_{br}Z_{eff}\sqrt{T}},
$$
with $\left<\sigma_{fusion}u\right>$ the fusion reactivity (which is also a function of $T$, approximately of $T^2$ for the temperatures of interest here), $\epsilon_\alpha=3.52\,\mathrm{MeV}$ the energy released by the $\alpha$-particles (assuming here that we have a DT-fusion reaction), $c_{br}\approx1.04\cdot 10^{-19}\,\mathrm{m}^3\mathrm{s}^{-1}\sqrt{\mathrm{eV}}$, $Z_{eff}$ the effective charge number which we assume to be 1 here. The left term in the denominator describes the heating by the $\alpha$-particles produced in the DT fusion reaction and the right term the losses due to bremsstrahlung.
The fusion triple product is often plotted as a function of the temperature, as shown in the following plot.

Some more interesting plots are shown for example here or here or here. I have chosen those particular examples because they include a limit for bremsstrahlung losses (the linear function on the top left). 
My question is how to calculate this limit such that it can be included in those plots?
(I have the feeling that the answer is quite obvious but I can't get it at the moment...)
 A: The above definition of the Lawson criterion derives from the following inequality
\begin{equation}
P_f \geq \frac{W}{\tau'_E} + P_{rad}\,,
\end{equation}
where $P_f= \frac{1}{4} n^2 \langle \sigma v\rangle \epsilon_{\alpha}$ is the fusion power density, $W=3 n T$ is the total energy density of the D-T plasma, $\tau'_E$ is an energy confinement time and $P_{rad}=c_{br} n^2 \sqrt{T}$ is the power density loss due to Bremsstrahlung. Solving for $n \tau'_E$ and multiplying by $T$ gives the following lower limit on the triple product
\begin{equation}
n T \tau'_E\geq \frac{12 T^2}{\langle \sigma v\rangle \epsilon_{\alpha}-4 c_{br}\sqrt{T}}\,.
\end{equation}
I suspect that the plots in the links above use a different definition of the Lawson criterion,
\begin{equation}
P_f \geq \frac{W}{\tau_E}\,,
\end{equation}
where $\tau_E$ is the total energy confinement time (including radiative losses).
Re-arranging again gives a different lower limit on the triple product,
\begin{equation}
n T \tau_E\geq \frac{12 T^2}{\langle \sigma v\rangle \epsilon_{\alpha}}\,.
\end{equation}
Now, by our definition we also require that the power density loss due to Bremsstrahlung is less than the total energy loss rate per unit volume,
\begin{equation}
P_{rad} \leq \frac{W}{\tau_E}\,,
\end{equation}
Re-arranging gives the following upper limit on the triple product
\begin{equation}
n T \tau_E \leq \frac{3 T^{3/2}}{c_{br}}\,.
\end{equation}
Using reactivities $\langle \sigma v\rangle$ from Table VII in a paper by Bosch and Hale (1992), I have plotted the lower and upper limits on the triple product (see below). 

