Let $a_1,\ldots,a_n$ be points on the unit circle. Let $P(z)=(z-a_1)\cdots(z-a_n)$. Prove that there exists a point $b$ on the unit circle such that $|P(b)|=1$.
My solution: $|P(0)|=1$, and $P$ is holomorphic in $|z|\leq 1$. By the maximum principle, there exists $z$ on the unit circle such that $|P(z)|>1$. But $P(a_1)=0$. By intermediate value theorem, there exists $b$ on the unit circle such that $|P(b)|=1$.
Question: Is there a way to avoid the maximum principle?
We can use Rouché's theorem to avoid the maximum principle ;)
Suppose there were no $b$ on the unit circle with $\lvert P(b)\rvert = 1$. Since $P$ has zeros on the unit circle, we would then have $\lvert P(z)\rvert < 1$ for all $z \in \mathbb{S}$.
Consider $f(z) = (-1)^n \prod\limits_{k=1}^n a_k$ and $g(z) = f(z) - P(z)$. Since
$$\lvert f(z) - g(z)\rvert = \lvert P(z)\rvert < 1 = \lvert f(z)\rvert$$
on $\mathbb{S}$, Rouché's theorem asserts that $f$ and $g$ have the same number of zeros in the unit disk. But $f$ is constant and nonzero, and $g(0) = f(0) - P(0) = 0$, contradiction.
Earnestly: the maximum principle provides a quick and direct natural proof, use it.
