Total-effort formula for a proportional contest with outside effort ($H_k(R+\delta {)}^2-(k-1)(R+\delta )-\delta =0$)

With $k>0$ active players of harmonic mean valuation ${\bar{v}}_k$, unit-cost Nash equilibrium effort satisfies $R+\delta =\frac{{\bar{v}}_k}{2k}[(k-1)+\sqrt{(k-1{)}^2+4k\delta /{\bar{v}}_k}]$. Sum the active-player first-order equations $b_i=(R+\delta )(1-(R+\delta )/v_i)$ to obtain the quadratic. If $\delta$ is at least the largest valuation, no one is active and $R=0$. With no outside effort, at least two players must be active.