Harvard-MIT Mathematics Tournament

Solution:
Answer: $\sqrt{3} \sqrt[3]{2}$

Write $a_{1}$ in its polar form $r e^{i \theta}$ where $0 \leq \theta \leq \frac{\pi}{2}$. Suppose $a_{2}, a_{3}$ and $r$ are fixed so that the denominator is constant. Write $a_{2}+a_{3}$ as $s e^{i \phi}$. Since $a_{2}$ and $a_{3}$ have non-negative real and imaginary parts, the angle $\phi$ lies between $0$ and $\frac{\pi}{2}$. Consider the function
$$f(\theta)=\left|a_{1}+a_{2}+a_{3}\right|^{2}=\left|r e^{i \theta}+s e^{i \phi}\right|^{2}=r^{2}+2 r s \cos (\theta-\phi)+s^{2}$$
Its second derivative is $f''(\theta)=-2 r s (\cos (\theta-\phi))$. Since $-\frac{\pi}{2} \leq (\theta-\phi) \leq \frac{\pi}{2}$, we know that $f''(\theta)<0$ and $f$ is concave. Therefore, to minimize $f$, the angle $\theta$ must be either $0$ or $\frac{\pi}{2}$. Similarly, each of $a_{1}, a_{2}$ and $a_{3}$ must be either purely real or purely imaginary to minimize $f$ and the original fraction.

By the AM-GM inequality, if $a_{1}, a_{2}$ and $a_{3}$ are all real or all imaginary, then the minimum value of the fraction is $3$. Now suppose only two of the $a_{i}$'s, say, $a_{1}$ and $a_{2}$ are real. Since the fraction is homogenous, we may fix $a_{1}+a_{2}$—let the sum be $2$. The term $a_{1} a_{2}$ in the denominator achieves its maximum only when $a_{1}$ and $a_{2}$ are equal, i.e. when $a_{1}=a_{2}=1$. Then, if $a_{3}=k i$ for some real number $k$, then the expression equals
$$\frac{\sqrt{k^{2}+4}}{\sqrt[3]{k}}$$
Squaring and taking the derivative, we find that the minimum value of the fraction is $\sqrt{3} \sqrt[3]{2}$, attained when $k=\sqrt{2}$. With similar reasoning, the case where only one of the $a_{i}$'s is real yields the same minimum value.