Problem 1584

1. Initial Considerations:
- We are given 1400 real numbers and need to prove that we can choose three of them, say $x, y, z$, such that:
$$\left|\frac{(x-y)(y-z)(z-x)}{x^4+y^4+z^4+1}\right| < 0.009$$
- Denote the given expression by $f(x, y, z)$ for brevity.

2. Case Analysis:
- Case 1: If $b \geq 500$:
$$f(b, y, z) < \frac{1}{4b} \leq \frac{1}{4 \times 500} = \frac{1}{2000} = 0.0005$$
Since $0.0005 < 0.009$, this case is satisfied.

- Case 2: If $b < 500$:
- Sub-case 2.1: If $b \geq 160$:
- Consider two sub-cases:
- Sub-case 2.1.1: There are two other numbers $y > z$ in the interval $\left[\frac{8}{25}b, b\right]$:
$$f(b, y, z) < \frac{\frac{1}{4} \cdot \left(\frac{17b}{25}\right)^3}{b^4 + \left(\frac{8}{25}b\right)^4 + \left(\frac{8}{25}b\right)^4} = \frac{122825}{1595268b} < \frac{122825}{1595268 \times 160} \approx 0.0004812$$
Since $0.0004812 < 0.009$, this sub-case is satisfied.
- Sub-case 2.1.2: At least 699 numbers belong to the interval $\left[0, \frac{8}{25}b\right)$:
- The length of this interval is $\frac{8}{25}b < 160$.
- Thus, we can always suppose that at least 698 numbers $a_1, \dots, a_{698}$ belong to $[0, 160)$.

- Sub-case 2.2: If $b < 160$:
- Consider 160 intervals $[0, 1), [1, 2), \dots, [159, 160)$.
- If there are 5 numbers in some interval $[k, k + 1)$, $k \geq 3$, then at least 3 numbers lie in either $\left[k, k + \frac{1}{2}\right)$ or $\left[k + \frac{1}{2}, k + 1\right)$:
$$f(x, y, z) \leq \frac{\frac{1}{4} \cdot \left(\frac{1}{2}\right)^3}{3k^4 + 1} \leq \frac{\frac{1}{4} \cdot \left(\frac{1}{2}\right)^3}{3 \cdot 3^4 + 1} \approx 0.000128$$
Since $0.000128 < 0.009$, this sub-case is satisfied.

- Otherwise, in each interval $[k, k + 1)$, $3 \leq k \leq 159$, there are at most 4 numbers. So, there are at least $698 - 157 \times 4 = 70$ numbers distributed among the three intervals $[0, 1), [1, 2), [2, 3)$. Thus, there is an interval containing at least 24 numbers, from which we can choose three $x > y > z$ belonging to a sub-interval of length $\frac{1}{8}$:
$$f(x, y, z) < \frac{1}{4} \cdot \left(\frac{1}{8}\right)^3 \approx 0.000488$$
Since $0.000488 < 0.009$, this sub-case is satisfied.

3. Conclusion:
- In all cases, we have shown that there exist three numbers $x, y, z$ such that:
$$\left|\frac{(x-y)(y-z)(z-x)}{x^4+y^4+z^4+1}\right| < 0.009$$

$\blacksquare$