Proof. We prove the result in the special case of just two miners, since the arbitrary case is a straightforward generalization. Let their compute powers be $\hat{C}_1$ and $\hat{C}_2$, respectively. We have that $$\mathbb{P}[T_\min < t] = 1 - \mathbb{P}[T_1 \geq t\ \&\ T_2 \geq t]$$

Assuming the two times are independent we have $$\begin{eqnarray*} \mathbb{P}[T_1 \geq t\ \&\ T_2 \geq t] & = & \mathbb{P}[T_1 \geq t]\mathbb{P}[ T_2 \geq t] \\ & = & (1 - 1/D)^{x / (S \hat{C}_{1})}(1 - 1/D)^{x / (S \hat{C}_{2})} \\ & = & (1 - 1/D)^{(x / S) (1/\hat{C}_{1} + 1/\hat{C}_{2})} \\ \end{eqnarray*}$$

We want $1/\hat{C}_{pooled} = 1/\hat{C}_{1} + 1/\hat{C}_{2}$, hence $$\hat{C}_{pooled} = (\hat{C}_{1}^{-1} + \hat{C}_{2}^{-1})^{-1}$$

$\Box$