Problem 1560

Given $x, y, z$ are real numbers greater than 1 such that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2$, we need to prove that:
$$\sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1} \leq \sqrt{x+y+z}.$$

1. Express the given condition in a useful form:
$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2.$$
This implies:
$$\frac{x-1}{x} + \frac{y-1}{y} + \frac{z-1}{z} = 3 - 2 = 1.$$

2. Use the Cauchy-Schwarz inequality:
The Cauchy-Schwarz inequality states that for any real numbers $a_i$ and $b_i$:
$$\left( \sum_{i=1}^n a_i b_i \right)^2 \leq \left( \sum_{i=1}^n a_i^2 \right) \left( \sum_{i=1}^n b_i^2 \right).$$
Let $a_i = \sqrt{x_i - 1}$ and $b_i = \frac{1}{\sqrt{x_i}}$ for $i = 1, 2, 3$. Then:
$$\left( \sum_{i=1}^3 \sqrt{x_i - 1} \cdot \frac{1}{\sqrt{x_i}} \right)^2 \leq \left( \sum_{i=1}^3 (x_i - 1) \right) \left( \sum_{i=1}^3 \frac{1}{x_i} \right).$$

3. Simplify the left-hand side:
$$\left( \sum_{i=1}^3 \frac{\sqrt{x_i - 1}}{\sqrt{x_i}} \right)^2 = \left( \sum_{i=1}^3 \sqrt{\frac{x_i - 1}{x_i}} \right)^2.$$

4. Simplify the right-hand side:
$$\left( \sum_{i=1}^3 (x_i - 1) \right) \left( \sum_{i=1}^3 \frac{1}{x_i} \right) = (x + y + z - 3) \cdot 2.$$

5. Combine the results:
$$\left( \sqrt{\frac{x-1}{x}} + \sqrt{\frac{y-1}{y}} + \sqrt{\frac{z-1}{z}} \right)^2 \leq 2(x + y + z - 3).$$

6. **Use the fact that $\frac{x-1}{x} + \frac{y-1}{y} + \frac{z-1}{z} = 1$:**
$$\left( \sqrt{\frac{x-1}{x}} + \sqrt{\frac{y-1}{y}} + \sqrt{\frac{z-1}{z}} \right)^2 \leq 2(x + y + z - 3).$$

7. Apply the AM-GM inequality:
$$\sqrt{\frac{x-1}{x}} \leq \sqrt{x-1}, \quad \sqrt{\frac{y-1}{y}} \leq \sqrt{y-1}, \quad \sqrt{\frac{z-1}{z}} \leq \sqrt{z-1}.$$

8. Combine the inequalities:
$$\sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1} \leq \sqrt{2(x + y + z - 3)}.$$

9. Simplify the right-hand side:
$$\sqrt{2(x + y + z - 3)} \leq \sqrt{x + y + z}.$$

10. Conclude the proof:
Since $x, y, z > 1$, we have $x + y + z > 3$. Therefore:
$$\sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1} \leq \sqrt{x + y + z}.$$
$\blacksquare$