Problem 1440

To find the smallest real number $C$ such that for any positive integers $x \neq y$, the following inequality holds:
$$\min(\{\sqrt{x^2 + 2y}\}, \{\sqrt{y^2 + 2x}\}) < C,$$
we need to analyze the behavior of the fractional parts of the expressions $\sqrt{x^2 + 2y}$ and $\sqrt{y^2 + 2x}$.

1. **Identify the best possible constant $C$:**
The best possible constant is $\varphi - 1 = \frac{1}{\varphi} = \frac{\sqrt{5} - 1}{2} \approx 0.618$, where $\varphi$ is the golden ratio, $\varphi = \frac{1 + \sqrt{5}}{2}$.

2. Construct a counterexample:
Consider large $x$ and $y$ such that $y = \lfloor \varphi x \rfloor$. This ensures $x \neq y$. We then have:
$$\sqrt{x^2 + 2y} = \sqrt{x^2 + 2\varphi x + \mathcal{O}(1)} = x + \varphi + o(1),$$
and
$$\sqrt{y^2 + 2x} = \sqrt{y^2 + \frac{2y}{\varphi} + \mathcal{O}(1)} = y + \frac{1}{\varphi} + o(1).$$
Since $\{\varphi\} = \left\{\frac{1}{\varphi}\right\} = \varphi - 1$, this construction shows that for any $C > \varphi - 1$ and sufficiently large $x$ and $y$, the inequality does not hold.

3. **Prove that $C = \varphi - 1$ works:**
Assume for contradiction that $x \neq y$ satisfies $\{\sqrt{x^2 + 2y}\} \ge C$ and $\{\sqrt{y^2 + 2x}\} \ge C$. Without loss of generality, assume $x < y$.

- From $\{\sqrt{y^2 + 2x}\} \ge C$, we get:
$$\sqrt{y^2 + 2x} \ge y + C.$$
Squaring both sides:
$$y^2 + 2x \ge y^2 + 2Cy + C^2,$$
which simplifies to:
$$2x \ge 2Cy + C^2 \implies x \ge Cy + \frac{C^2}{2}.$$

- Since $y > x$, we have $x^2 + 2y > x + 1$. From $\{\sqrt{x^2 + 2y}\} \ge C$, we get:
$$\sqrt{x^2 + 2y} \ge x + C + 1.$$
Squaring both sides:
$$x^2 + 2y \ge x^2 + 2Cx + C^2 + 2x + 1,$$
which simplifies to:
$$2y \ge 2Cx + C^2 + 2x + 1 \implies y \ge (C + 1)x + \frac{C^2 + 1}{2}.$$

- Since $C + 1 = \frac{1}{C}$, the inequalities $x \ge Cy + \frac{C^2}{2}$ and $y \ge (C + 1)x + \frac{C^2 + 1}{2}$ contradict each other for $C = \varphi - 1$.

Thus, the smallest real number $C$ that satisfies the given condition is $\varphi - 1$.

The final answer is $\boxed{\varphi - 1}$.