Problem 1556

1. Substitution and Simplification:
Let $z = 1 - x - y$. Then, we have $x + y + z = 1$. We need to prove:
$$\frac{x^2}{x+y} + \frac{y^2}{1-x} + \frac{(1-x-y)^2}{1-y} \geq \frac{1}{2}.$$
Substituting $z$ into the expression, we get:
$$\frac{x^2}{1-z} + \frac{y^2}{1-y} + \frac{z^2}{1-z}.$$

2. Rearrangement Inequality:
Suppose $z$ is negative. Then $|z| = |1 - x - y| < x, y$ since $1 - x, 1 - y > 0$. Thus, $z^2 < x^2, y^2$. Without loss of generality, assume $z < x < y$. If $z$ is positive, the same argument holds since $x, y, z$ are symmetric.

3. Applying Rearrangement Inequality:
By the rearrangement inequality, the sum is less than or equal to:
$$\frac{x^2}{1-x} + \frac{y^2}{1-y} + \frac{z^2}{1-z}.$$

4. Derivative and Jensen's Inequality:
Consider the function $f(t) = \frac{t^2}{1-t}$. The derivative is:
$$f'(t) = \frac{d}{dt} \left( \frac{t^2}{1-t} \right) = \frac{2t(1-t) + t^2}{(1-t)^2} = \frac{t(2-t)}{(1-t)^2}.$$
This derivative is strictly increasing in the interval $(0, 1)$.

5. Applying Jensen's Inequality:
By Jensen's inequality for the convex function $f(t)$ over the interval $(0, 1)$, we have:
$$f(x) + f(y) + f(z) \geq 3f\left( \frac{x+y+z}{3} \right).$$
Since $x + y + z = 1$, we get:
$$f\left( \frac{1}{3} \right) = \frac{\left( \frac{1}{3} \right)^2}{1 - \frac{1}{3}} = \frac{\frac{1}{9}}{\frac{2}{3}} = \frac{1}{6}.$$
Therefore:
$$\frac{x^2}{1-x} + \frac{y^2}{1-y} + \frac{z^2}{1-z} \geq 3 \cdot \frac{1}{6} = \frac{1}{2}.$$

6. Conclusion:
Thus, we have shown that:
$$\frac{x^2}{x+y} + \frac{y^2}{1-x} + \frac{(1-x-y)^2}{1-y} \geq \frac{1}{2}.$$
The equality holds when $x = y = z = \frac{1}{3}$.

$\blacksquare$