Problem 1423

1. Define the sides and diagonals:
Let $AB = a$, $BC = b$, $CD = c$, $AC = p$, and $BD = q$. Without loss of generality, assume $BC = 1$.

2. **Apply the Law of Cosines to $\triangle ABC$:**
Since $\angle B > 120^\circ$, we have:
$$p^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle B)$$
Given $\angle B > 120^\circ$, $\cos(\angle B) < -\frac{1}{2}$. Therefore:
$$p^2 \geq a^2 + 1 + 2a \cdot \frac{1}{2} = a^2 + 1 + a = a^2 + a + 1$$
$$p^2 > (a + \frac{1}{2})^2$$
Taking the square root on both sides:
$$p > a + \frac{1}{2}$$

3. **Apply the Law of Cosines to $\triangle BCD$:**
Similarly, since $\angle C > 120^\circ$, we have:
$$q^2 = BD^2 + BC^2 - 2 \cdot BD \cdot BC \cdot \cos(\angle C)$$
Given $\angle C > 120^\circ$, $\cos(\angle C) < -\frac{1}{2}$. Therefore:
$$q^2 \geq c^2 + 1 + 2c \cdot \frac{1}{2} = c^2 + 1 + c = c^2 + c + 1$$
$$q^2 > (c + \frac{1}{2})^2$$
Taking the square root on both sides:
$$q > c + \frac{1}{2}$$

4. Combine the inequalities:
Adding the inequalities for $p$ and $q$:
$$p + q > (a + \frac{1}{2}) + (c + \frac{1}{2}) = a + c + 1$$

5. Relate to the original inequality:
Since $BC = 1$, we have:
$$p + q > a + b + c$$
Therefore:
$$|AC| + |BD| > |AB| + |BC| + |CD|$$

$\blacksquare$

The final answer is $\boxed{ |AC| + |BD| > |AB| + |BC| + |CD| }$