Find the smallest real number $C$, such that for any positive integers $x \neq y$ holds the following:

$$\min(\{\sqrt{x^2 + 2y}\}, \{\sqrt{y^2 + 2x}\})<C$$

Here $\{x\}$ denotes the fractional part of $x$. For example, $\{3.14\} = 0.14$.

[i]Proposed by Anton Trygub[/i]
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 \(\boxed{\varphi - 1}\).