Harmonics functions take minimum value on the boundary Let $u$ be harmonic on the bounded region $A$ and continuous on the closure of $A$. Then $u$ takes its minimum only on $\partial A$ unless $u$ is a constant.

How can we prove this result? Do we have to use the Maximum modulus principles? in my notes, the proof only says that considering $v = - u$, the conclusion follows. But, how? I dont see it.
$-u$ is a perfectly good harmonic function, so it obeys the mean value property. Also, it is clear that $-u$ attains a maximum when $u$ is at a minimum.

$\overline{A}$ is compact, so $v$ obtains a maximum on $\overline{A}$. Suppose $v:=-u$ attains this maximum at some point $z_0\in A^{\circ}$ (the interior of $A$). If $\varepsilon&gt;0$ is small enough that $D(z_0,\varepsilon)\subset A$ (this is possible because $z_0$ is in the interior), then the mean value property dictates that

$$
v(z_0)=\frac{1}{2\pi}\int_{0}^{2\pi}v(z_0+\varepsilon e^{i\theta})\,d\theta.
$$

Now, the expression on the right-hand side is precisely the average value of $v$ over $\partial D(z_0,\varepsilon)$. $v(z_0)$ is assumed to be a local maximum, so it cannot possibly be the average value of other points in a neighborhood of it unless it is constant. Therefore $v$ is constant on $D(z_0,\varepsilon)$.

We just need to show that this extends to mean $v$ is globally constant. I believe this argument should work:

Let $z_1\in A^\circ$ be distinct from $z_0$. Then there exists some continuous curve $\gamma:(0,1)\to A^{\circ}$ such that $\gamma(0)=z_0$ and $\gamma(1)=z_1$ (here I'm assuming $A$ is connected--else the statement is not true). Let $m$ be the infimum of $\text{dist}(\gamma,\partial A)$. Since the image of $\gamma$ is compact, there exists a finite sequence of overlapping open disks $\{U_i\}$ with radius $\min\!\big(m/2,\varepsilon/2\big)$ such that $U_i\subset A^{\circ}$ and $\bigcup_iU_i$ covers the image of $\gamma$. If $U_0$ contains $z_0$, then $U_0\subset D(0,\varepsilon)$ and $v$ is constant on $U_0$. There is a further disk $U_1$ such that $U_0\cap U_1\neq\emptyset$ and, since $U_1$ has radius at most $\varepsilon/2$, the center of $U_1$ attains the same maximum. By the same argument $v$ is constant on $U_1$ as well. This process can be continued to show that $v$ is constant on every $U_i$, and hence is constant in a neighborhood of $z_1$. $z_1$ was chosen arbitrarily, so $v$ is constant on all of $A$.

The argument got a little technical, but the main idea is you take a finite sequence of disks--the first containing $z_0$ and the last containing $z_1$--each small enough so that they are all contained within $A$ and such that the center of each disk is contained within the previous disk (that way after showing $v$ is constant on one disk, it attains the maximum at the center of the subsequent disk, so the previous argument will immediately apply to show it is constant on that disk as well).
One could prove the minimum modulus principle via complex analysis and knowing it for holomorphic functions. I assume you realize the maximum modulus principle for holomorphic functions (or , if not, you can easily prove it knowing it for harmonic functions)
In order to prove this you should look at $g=\dfrac{1}{f}$ which is also holomorphic (since we assume $f\ne 0$) and use the maximum modulus principle.

Edit: I would take the downvotes as a hint for lack of explaination:
Harmonic functions can be viewed as a real or imaginary part of holomorphic functions. The maximum (minimum) modulus principles of holomorphic functions can be used to prove the same principles for harmonic functions.