= Solution
The <Lagrange sufficiency theorem> says that if a feasible point $x^*$ and nonnegative multipliers satisfy complementary slackness and $x^*$ globally maximizes the <Lagrangian function in constrained optimization>, then $x^*$ globally solves the constrained maximization problem. Indeed, for every feasible $x$, the signs of the constraints give $f(x)\leq L(x)\leq L(x^*)=f(x^*)$. Concavity and the <Karush-Kuhn-Tucker conditions> are a common way to verify the global Lagrangian maximum.
For fixed $y$, A maximizes
$$
F(x,y)=\sum_i u_i\frac{x_i-y_i}{x_i+y_i}
=\sum_i u_i-2\sum_i\frac{u_iy_i}{x_i+y_i}
$$
subject to $x_i\geq0$ and $\sum_i x_i=X$. This is a <concave function> of $x$. On every active coordinate, the multiplier equation is
$$
\frac{2u_iy_i}{(x_i+y_i)^2}=\lambda.
$$
Thus A's optimal allocation is given by the <water-filling algorithm>:
$$
\boxed{x_i=\left(\sqrt{\frac{2u_iy_i}{\lambda}}-y_i\right)_+,}
$$
where $\lambda>0$ is chosen so that $\sum_i x_i=X$. Coordinates with zero expression receive no resource. This is a <water-filling algorithm>. The formula gives the optimum when every $y_i>0$. If $y_i=0$, any $x_i>0$ already gives A the whole market $u_i$ in that region, so A assigns such coordinates arbitrarily small positive amounts and applies water filling to the remaining budget; without a convention for $0/0$, the resulting value can be a supremum rather than an attained maximum.
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