China Team Selection Test

We aim to find the smallest positive number $\lambda$ such that for any complex numbers $z_1, z_2, z_3 \in \{z \in \mathbb{C} \mid |z| < 1\}$ with $z_1 + z_2 + z_3 = 0$, the following inequality holds:
$$\left|z_1z_2 + z_2z_3 + z_3z_1\right|^2 + \left|z_1z_2z_3\right|^2 < \lambda.$$

First, we show that $\lambda \geq 1$. Consider $z_1 = 1 - \epsilon$, $z_2 = 0$, and $z_3 = \epsilon - 1$, where $\epsilon$ is a small positive real number. Then,
$$z_1 + z_2 + z_3 = (1 - \epsilon) + 0 + (\epsilon - 1) = 0.$$
We have
$$|z_1z_2 + z_2z_3 + z_3z_1|^2 = |(1 - \epsilon)(\epsilon - 1)|^2 = (1 - \epsilon^2)^2,$$
which can be made arbitrarily close to 1 as $\epsilon \to 0$. Hence, $\lambda \geq 1$.

Now, we prove that $\lambda = 1$ works. Let $z_k = r_k (\cos \theta_k + i \sin \theta_k)$ for $k = 1, 2, 3$. Given $z_1 + z_2 + z_3 = 0$, we have:
$$\sum_{k=1}^3 r_k \cos \theta_k = 0 \quad \text{and} \quad \sum_{k=1}^3 r_k \sin \theta_k = 0.$$

Squaring and adding these equations, we get:
$$r_1^2 + r_2^2 + 2r_1r_2 \cos(\theta_2 - \theta_1) = r_3^2.$$

Thus,
$$\cos(\theta_2 - \theta_1) = \frac{r_3^2 - r_1^2 - r_2^2}{2r_1r_2}.$$

We then have:
$$2r_1^2r_2^2 \cos(2\theta_2 - 2\theta_1) = 2r_1^2r_2^2 (2 \cos^2(\theta_2 - \theta_1) - 1) = (r_3^2 - r_1^2 - r_2^2)^2 - 2r_1^2r_2^2 = r_1^4 + r_2^4 + r_3^4 - 2r_1^2r_3^2 - 2r_2^2r_3^2.$$

Adding cyclic permutations, we get:
$$\sum_{1 \leq i < j \leq 3} 2r_i^2r_j^2 \cos(2\theta_j - 2\theta_i) = 3(r_1^4 + r_2^4 + r_3^4) - 4(r_1^2r_2^2 + r_2^2r_3^2 + r_3^2r_1^2).$$

Given $z_1 + z_2 + z_3 = 0$, we can swap $z_1z_2 + z_2z_3 + z_3z_1$ with $\frac{1}{2}(z_1^2 + z_2^2 + z_3^2)$. Thus,
$$\left|z_1z_2 + z_2z_3 + z_3z_1\right|^2 + \left|z_1z_2z_3\right|^2 = \frac{1}{4} \left|z_1^2 + z_2^2 + z_3^2\right|^2 + |z_1z_2z_3|^2.$$

This simplifies to:
$$\frac{1}{4} \left( (\sum r_i^2 \cos 2\theta_i)^2 + (\sum r_i^2 \sin 2\theta_i)^2 \right) + r_1^2 r_2^2 r_3^2.$$

Using the identities and properties of trigonometric functions and binomial coefficients, we get:
$$\frac{1}{4} \left( r_1^4 + r_2^4 + r_3^4 + 2 \sum_{1 \leq i < j \leq 3} r_i^2 r_j^2 \cos(2\theta_j - 2\theta_i) \right) + r_1^2 r_2^2 r_3^2.$$

This reduces to:
$$r_1^4 + r_2^4 + r_3^4 - (r_1^2 r_2^2 + r_2^2 r_3^2 + r_3^2 r_1^2) + r_1^2 r_2^2 r_3^2 \leq 1 - (1 - r_1^2)(1 - r_2^2)(1 - r_3^2) \leq 1.$$

Thus, $\lambda = 1$ works. Therefore, the smallest positive number $\lambda$ is:
$$\boxed{1}.$$