= Solution
First exclude boundary profiles. At $(0,0)$, player $i$ can replace its payoff $v_i/2$ by $v_i-\varepsilon^2>v_i/2$ using a sufficiently small positive effort. If one effort is positive and the other is zero, the positive bidder can lower its effort while retaining the entire prize. Thus a pure <Nash equilibrium> of this <proportional allocation contest> must have both efforts positive.
Against $b_j>0$, player $i$'s payoff is a <strictly concave function> of $b_i\geq0$, since
$$
\frac{\partial^2 s_i}{\partial b_i^2}
=-\frac{2v_i b_j}{(b_i+b_j)^3}-2<0.
$$
Its derivative at zero is positive and its payoff tends to negative infinity as its own effort tends to infinity. Hence its unique <best response> is the positive solution of the <first-order condition>. At an equilibrium, writing $B=b_1+b_2$, these conditions are
$$
\frac{v_1b_2}{B^2}=2b_1,\qquad
\frac{v_2b_1}{B^2}=2b_2.
$$
Dividing them gives $b_1/b_2=\sqrt{v_1/v_2}$, and multiplying them gives $B^4=v_1v_2/4$. Therefore the <quadratic-cost two-player proportional contest> has
$$
\boxed{B=\frac{(v_1v_2)^{1/4}}{\sqrt2},\qquad
b_i^*=\frac{\sqrt{v_i}}{\sqrt{v_1}+\sqrt{v_2}}\,
\frac{(v_1v_2)^{1/4}}{\sqrt2}.}
$$
Both efforts are positive, and <strict concavity> makes them global <best responses>. The first-order conditions have only this positive solution, while the boundary profiles have already been excluded. This proves both existence and uniqueness. The resulting <winning probabilities> are $x_i^*=\sqrt{v_i}/(\sqrt{v_1}+\sqrt{v_2})$; for equal values $v$, each effort is $\sqrt{v/8}$.
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