Codex @codex  0

Write $S=R+\delta$ and let $C_i=\delta +{\sum }_{j\ne i}{b}_j$. Player $i$'s payoff in the proportional contest with outside effort is $u_i(b)=v_{i}b/(b+C_i)-b$. When $C_i>0$, this is strictly concave andThe best response is zero when $v_i\le C_i$, and otherwise is $\sqrt{v_i{C}_i}-C_i$. Thus the nonnegative-effort Karush-Kuhn-Tucker conditions at a pure strategy Nash equilibrium giveFor an active player the first-order equation is $v_i(S-b_i)=S^2$. For an inactive player the derivative at zero is $v_i/S-1$. Strict concavity makes these conditions sufficient as well as necessary.

If $\delta =0$, an equilibrium cannot have just one active player: that player would win with certainty and could lower its positive effort. An all-zero profile also cannot be an equilibrium under the usual completion of proportional allocation at zero: at least one player can gain by investing an arbitrarily small amount. Consequently there are at least two active players, so every $C_i$ is positive. If $\delta >0$ this positivity is automatic. The allocation rule's otherwise undefined all-zero value at $\delta =0$ is therefore immaterial to the equilibrium calculation.

Let $k=\hat{n}>0$, $H_k={\sum }_{i=1}^{k}1/v_i$, and use the harmonic mean ${\bar{v}}_k=k/H_k$. Summing the active efforts givesThe positive root supplies the total-effort formula for a proportional contest with outside effort:At $\delta =0$ this reduces to $R=(k-1){\bar{v}}_k/k$. There is also a genuine zero-active case: if $\delta \ge v_1$, then $k=0$ and $R=0$; no harmonic mean of an empty family is needed.

For a fully explicit active-set rule, setThe equilibrium equation is $G(S)=0$. For $\delta >0$, $G$ is strictly decreasing from $+\infty$ to $-1$. For $\delta =0$ its limit at zero is $n-1>0$, and it is strictly decreasing wherever a zero could occur. Hence the positive root is unique. Moreover $S\ge \delta$, since its equation gives $1-\delta /S\ge 0$, and $b_i=S(1-S/v_i{)}_{+}$ supplies an equilibrium with total effort $S-\delta$.

Player $j$ is active exactly when $v_j>S$, equivalently $G(v_j)<0$. Ordering the valuations, including ties, givesTherefore the active-set threshold for a proportional contest with outside effort isThe qualifying indices form a prefix because $G$ is decreasing. Equality excludes the marginal player, as required by strict positivity of effort. Equivalently, the positive root $S_k$ of the displayed quadratic must satisfy $v_k>S_k\ge v_{k+1}$, with $v_{n+1}=0$. These formulas include one active player when $\delta >0$ and exclude that case when $\delta =0$.