= Solution
Let $w_m=w(m/n)$, so a type $v$ values a prize at $v w_m$. In a symmetric increasing <Bayes-Nash equilibrium> of this <rank-order contest>, a player reporting type $t$ wins when at most $m-1$ opponents have larger types. Its <winning probability> is
$$
q_m(t)=\sum_{j=0}^{m-1}\binom{n-1}{j}(1-t)^j t^{n-1-j}.
$$
Here the count of opponents above $t$ has a <binomial distribution>. Differentiation, or the associated <order statistic> density, gives
$$
q_m'(t)=\frac{(n-1)!}{(n-m-1)!(m-1)!}\,t^{n-m-1}(1-t)^{m-1}.
$$
Define the effort by the <all-pay effort identity>
$$
\boxed{b(v)=w_m\int_0^v t q_m'(t)\,dt.}
$$
This also verifies equilibrium globally. The derivative of a type $v$'s payoff from reporting $t$ is $w_m(v-t)q_m'(t)$, positive before $v$ and negative after $v$. Thus truthful reporting is a <best response>. Bidding above the maximal equilibrium effort gains no additional <winning probability>. Type zero chooses zero effort.
By exchanging the two integrations, the <expected value> of total effort is
$$
\begin{aligned}
R_m&=n\int_0^1b(v)\,dv
=nw_m\int_0^1t(1-t)q_m'(t)\,dt\\
&=\boxed{\frac{m(n-m)}{n+1}\,w(m/n).}
\end{aligned}
$$
The last integral is the moment $\mathbb E[X(1-X)]$ for a <Beta distribution> with parameters $n-m$ and $m$, namely $m(n-m)/(n(n+1))$. This is the <uniform-value multi-prize all-pay effort formula>.
Put $x=m/n$ and $h(x)=w(x)x(1-x)$. The positive constant $n^2/(n+1)$ multiplying $h(x)$ does not affect the maximizing $m$. Since
$$
h'(x)=w'(x)x(1-x)+w(x)(1-2x),
$$
the assumed inequality makes $h$ nonincreasing on the feasible interval. Hence \b[one prize maximizes expected total effort].
For the power family, the PDF gives $w(x)=x^{-\alpha}$. Then
$$
h(x)=x^{1-\alpha}(1-x),\qquad
h'(x)=x^{-\alpha}\bigl[(1-\alpha)-(2-\alpha)x\bigr].
$$
If $\alpha\geq1$, this derivative is strictly negative for $0<x<1$: the bracket is affine and its values at the endpoints are $1-\alpha\leq0$ and $-1$. Thus \b[$m=1$ is optimal].
If $0<\alpha<1$, the unique continuous maximizer is
$$
\boxed{x_* =\frac{1-\alpha}{2-\alpha}.}
$$
The objective strictly increases before $x_*$ and decreases after it. Therefore its discrete maximizer lies among
$$
\boxed{\left\{\lfloor nx_*\rfloor,\ \lfloor nx_*\rfloor+1\right\}\cap\{1,\ldots,n-1\}.}
$$
Compare the surviving candidates using $m^{1-\alpha}(n-m)$, since $R_m=n^\alpha m^{1-\alpha}(n-m)/(n+1)$. If $nx_*$ is an integer, that integer is the unique maximizer; if its floor is zero, the only feasible candidate is $1$. This proves the <discrete prize-count optimization for a power-valued contest> including the endpoint cases.
The sufficient condition need only hold on $[1/n,(n-1)/n]$. The power family is undefined at zero, so the printed endpoint $x=0$ cannot apply to it. More generally, a positive finite value $w(0)$ would make the displayed inequality fail at zero. The design argument uses only positive feasible prize fractions and needs no value there.
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