To solve the inequality

\[
\sqrt{x-6} - \sqrt{10-x} \geq 1,
\]

follow these steps:

1. **Identify the domain**:
- For \(\sqrt{x-6}\) to be real, \(x - 6 \geq 0\) implies \(x \geq 6\).
- For \(\sqrt{10-x}\) to be real, \(10 - x \geq 0\) implies \(x \leq 10\).
- Thus, the domain is \(6 \leq x \leq 10\).

2. **Rearrange the inequality**:
\[
\sqrt{x-6} \geq \sqrt{10-x} + 1.
\]

3. **Square both sides** (noting that this can introduce extraneous solutions):
\[
x - 6 \geq (10 - x) + 2\sqrt{10 - x} + 1.
\]
Simplifying this gives:
\[
x - 6 \geq 11 - x + 2\sqrt{10 - x}.
\]
Rearranging further:
\[
2x - 17 \geq 2\sqrt{10 - x}.
\]

4. **Divide by 2**:
\[
x - \frac{17}{2} \geq \sqrt{10 - x}.
\]

5. **Square both sides again**:
\[
\left(x - \frac{17}{2}\right)^2 \geq 10 - x.
\]
Expanding the left side:
\[
x^2 - 17x + \frac{289}{4} \geq 10 - x.
\]
Rearranging gives:
\[
x^2 - 16x + \frac{249}{4} \geq 0.
\]

6. **Multiply through by 4** (to eliminate the fraction):
\[
4x^2 - 64x + 249 \geq 0.
\]

7. **Finding the roots** using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{64 \pm \sqrt{(-64)^2 - 4 \cdot 4 \cdot 249}}{2 \cdot 4}.
\]

Calculate the discriminant:
\[
4096 - 3968 = 128.
\]

The roots are:
\[
x = \frac{64 \pm \sqrt{128}}{8} = \frac{64 \pm 8\sqrt{2}}{8} = 8 \pm \sqrt{2}.
\]

8. **Test intervals**:
- The roots are approximately \(8 - \sqrt{2} \approx 6.586\) and \(8 + \sqrt{2} \approx 9.414\).
- Within the interval \([6, 10]\), the intervals to test are \([6, 8 - \sqrt{2})\), \((8 - \sqrt{2}, 8+\sqrt{2})\), and \((8 + \sqrt{2}, 10]\).

9. **Conclusion**:
- Check if the intervals satisfy the inequality. It turns out the inequality holds for \(x \in [8 - \sqrt{2}, 10]\) within the defined domain of \(6 \leq x \leq 10\).

Thus, the solution set is:

\[
[8 - \sqrt{2}, 10].
\]