3. Find all pairs of real numbers $(x ; y)$ that satisfy the inequality $\sqrt{x+y-1}+x^{4}+y^{4}-\frac{1}{8} \leq 0$. In your answer, write the greatest value of the product $xy$ for all found pairs $(x ; y)$.

(6 points)
Solution. Domain of definition: $x+y \geq 1$. We will prove the validity of the inequality $x^{4}+y^{4} \geq 1 / 8$ under this condition. For any $x$ and $y$, the inequality

$$
(x+y)^{2} \geq 4 x y \quad \Rightarrow \quad \frac{(x+y)^{2}}{2} \geq 2 x y
$$

holds. Then $x^{2}+y^{2}=(x+y)^{2}-2 x y \geq \frac{(x+y)^{2}}{2} \geq \frac{1}{2}, \quad x^{4}+y^{4} \geq \frac{\left(x^{2}+y^{2}\right)^{2}}{2} \geq \frac{1}{8}$.

The original inequality in the problem statement is satisfied under the condition
$\left\{\begin{array}{c}x+y=1, \\ x^{4}+y^{4}=1 / 8,\end{array} \Leftrightarrow\left\{\begin{array}{c}x^{2}+y^{2}=1-2 x y, \\ \left(x^{2}+y^{2}\right)^{2}-2(x y)^{2}=1 / 8,\end{array} \Rightarrow(1-2 x y)^{2}-2(x y)^{2}=1 / 8, \quad t=x y\right.\right.$,

$(1-2 t)^{2}-2 t^{2}=1 / 8, \quad 2 t^{2}-4 t+7 / 8=0, \quad t=7 / 4, t=1 / 4$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3205747dae2ad15e28c2g-02.jpg?height=154&width=1254&top_left_y=477&top_left_x=184)

Answer: 0.25.