An analytic function on a bounded domain $D$ attains its minimum on $\partial D$

Let $f(z)$ be an analytic function on a domain $D$ that has no zeros on $D$. Show that if $D$ is bounded, and if $f(z)$ extends continuously to the boundary $\partial D$ of $D$, then $|f(z)|$ attains its minimum on $\partial D$.

let $h(z)=\displaystyle \frac{1}{f(z)}$. Since $f(z)$ is analytic and nonzero on $D$ we have that $h(z)$ is analytic, hence harmonic on a bounded domain $D$. Also, since $f(z)$ extends continuously $D$ to $\partial D$ the so does $h(z)$. Now Since a continuous function on a compact set is bounded, we have that, $|h(z)|\le M$ for some $M$ and for all $z\in \partial D$. By Maximum Principle we have that $|h(z)|\le M$ for all $z\in D$. We need to check that $|f(z)|\not =0$ on $\partial D$.

I don't know how to use the fact that $f(z)$ has no zeros on D, to show that it has no zeros on $\partial D$

then we have $\frac{1}{M}\le |f(z)|$ for all $z\in \partial D$. And hence, $f(z)$ attains its minimum on $\partial D$.

If $f(z)=0$ for some $z\in\partial D$, then clearly $|f(z)|$ attains it's minimum (namely, zero) on the boundary. If $|f(z)|>0$ for all $z\in \partial D$, apply the maximum-modulus principle to $1/f$.