Problem 1448

1. We start with the given inequality:
$$\frac{1}{\frac{1}{a} + \frac{1}{b}} + \frac{1}{\frac{1}{c} + \frac{1}{d}} \le \frac{1}{\frac{1}{a+c} + \frac{1}{b+d}}$$

2. We use the formula for the harmonic mean:
$$\frac{1}{\frac{1}{x} + \frac{1}{y}} = \frac{xy}{x + y}$$
Applying this to our inequality, we get:
$$\frac{ab}{a + b} + \frac{cd}{c + d} \le \frac{(a+c)(b+d)}{a + b + c + d}$$

3. To prove this, we rewrite the inequality in a different form:
$$\frac{ab}{a + b} + \frac{cd}{c + d} \le \frac{(a+c)(b+d)}{a + b + c + d}$$

4. We introduce the following terms:
$$x = \frac{ab}{a + b}, \quad y = \frac{cd}{c + d}, \quad z = \frac{(a+c)(b+d)}{a + b + c + d}$$
So, we need to show:
$$x + y \le z$$

5. We use the identity:
$$\frac{a+b}{4} - \frac{ab}{a+b} + \frac{c+d}{4} - \frac{cd}{c+d} \ge \frac{a+b+c+d}{4} - \frac{(a+c)(b+d)}{a+b+c+d}$$

6. Simplifying the left-hand side:
$$\frac{(a-b)^2}{a+b} + \frac{(c-d)^2}{c+d}$$
and the right-hand side:
$$\frac{(a+c-b-d)^2}{a+b+c+d}$$

7. By the Cauchy-Schwarz inequality, we know:
$$\frac{(a-b)^2}{a+b} + \frac{(c-d)^2}{c+d} \ge \frac{(a+c-b-d)^2}{a+b+c+d}$$

8. Therefore, the inequality holds true:
$$\frac{ab}{a+b} + \frac{cd}{c+d} \le \frac{(a+c)(b+d)}{a+b+c+d}$$

$\blacksquare$