Just summing together all the comments and providing some more explicit calculations, we have conservation of four-momentum (which is the amalgamation of the conservation of energy and conservation of momentum), we have:
$$p^{\mu}{1}+p{2}^{\mu}=p_{1}'^{\mu}+p_{2}'^{\mu}+p_{\pi}^{\mu}$$
Taking the inner product of each side with itself, we get:
$$\left\langle p_{1}^{\mu}\middle|p_{1}^{\mu}\right\rangle+2\left\langle p_{1}^{\mu} \middle| p_{2}^{\mu}\right\rangle+\left\langle p_{2}^{\mu}\middle|p_{2}^{\mu}\right\rangle=\left\langle p_{1}'^{\mu} \middle| p_{1}'^{\mu}\right\rangle + 2\left\langle p_{1}'^\mu \middle| p_{2}'^{\mu} \right\rangle + 2\left\langle p_{1}'^{\mu} \middle| p_{\pi}^{\mu}\right\rangle + \left\langle p_{2}'^\mu \middle| p_{2}'^{\mu} \right\rangle + 2\left\langle p_{2}'^{\mu} \middle| p_{\pi}^{\mu} \right\rangle + \left\langle p_{\pi}^{\mu} \middle| p_{\pi}^{\mu}\right\rangle$$
We note that $p^{\mu}p'{\mu}=\left\langle p^{\mu} \middle| p'^{\mu}\right\rangle$ is invaraiant in all frames of reference and that $p^{\mu}p{\mu}=m^{2}c^{2}$, we can therefore simplify:
$$2m_{p}^{2}c^{2}+2\left\langle p_{1}^{\mu} \middle| p_{2}^\mu \right\rangle = 4m_{p}^{2}c^{2}+4m_{p}m_{\pi}c^{2}+m_{\pi}^{2}c^{2}$$
If we consider that the second proton is initially at rest we have: $p_{1}^{\mu}=\left(\frac{E}{c},\vec{p}\right)$ and therefore:
$$2m_{p}E=2m_{p}^{2}c^{2}+4m_{p}m_{\pi}c^{2}+m^{2}_{\pi}c^{2}$$
Rearranging we get:
$$E=m_{p}c^{2}+2m_{\pi}c^{2}+\frac{m_{\pi}^{2}c^{2}}{2m_{p}}$$
Using the constants $m_{p}=938\text{ GeV}/c^{2}$ and $m_{\pi}=139.6\text{ GeV}/c^{2}$, we get:
$$E=10.6653 \text{ TeV} \implies T = 10.6643 \text{ TeV}$$
So if a proton with 10 TeV of kinetic energy collided with a stationary proton, there is a chance that a pion will be produced. 66