Proof of maximal principle on Laplace Equation involving Poisson integral formula This question appeared on a past PDE exam I found while studying for my finals:

Let $u(r,\theta)$ be  solution to the Laplace equation in polar coordinates:
  $$u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta} = 0, \quad 0<r\le a$$
  which satisfies the boundary condition
  $$u(a,\theta)=f(\theta)$$
  where $u$ is given by Poisson's formula
  $$u(r,\theta)=\frac{a^2-r^2}{2\pi}\int_0^{2\pi} \frac{f(\xi)}{a^2-2ar\cos(\theta-\xi)+r^2 }\mathrm{d}\xi$$Prove that if $M=\max(|f(\theta)|)$, then $|u(r,\theta)| \le M$ for all $r \le a$ and all $\theta$.

This is obviously the maximal principle for harmonic functions and can be proved using analysis with balls and topological principles. However, none of this is covered in my class since its an undergraduate intro to PDE course.
The sample solution, given by the professor, is incorrect from what I can see and goes as follows:
\begin{align}
|u|&= \left|\frac{a^2-r^2}{2\pi}\int_0^{2\pi} \frac{f(\xi)}{a^2-2ar\cos(\theta-\xi)+r^2 }\mathrm{d}\xi\right|\\[1em]
 &= \frac{a^2-r^2}{2\pi} \left|\int_0^{2\pi} \frac{f(\xi)}{a^2-2ar\cos(\theta-\xi)+r^2 }\mathrm{d}\xi\right|\\[1em]
&\le \frac{a^2-r^2}{2\pi}\int_0^{2\pi} \left|\frac{f(\xi)}{a^2-2ar\cos(\theta-\xi)+r^2 }\right|\mathrm{d}\xi \\[1em]
&\le \frac{a^2-r^2}{2\pi}\int_0^{2\pi} \frac{|M|}{a^2-2ar+r^2 }\mathrm{d}\xi \\[1em]
&=\boxed{|M| \frac{a^2-r^2}{a^2-2ar+r^2}\frac{1}{2\pi} \int_0^{2\pi} \mathrm{d}\xi = |M| \frac{a-r}{a+r}} \le |M| \quad \blacksquare\\{}
\end{align}
The equality I have boxed is where the error arises since:
$$\frac{a^2-r^2}{a^2-2ar+r^2} = \frac{(a-r)(a+r)}{(a-r)^2} = \frac{a+r}{a-r} \ne \frac{a-r}{a+r}$$
by this logic the inequality is unbounded.
I have brought this up with the professor and he said that he would get back to me in a week as he agrees with me, but I honestly am too bugged by this to wait.
So what is the correct proof using the same method demonstrated by the professor? 
Clearly the error must arise from some mistake in the logic followed by the inequalities of absolute values. 
I tried to mend the solution by applying the triangle equality and/or reverse triangle inequality to the denominator of the integral but I can't arrive to the form
$$\frac{a^2-r^2}{a^2+2ar+r^2} = \frac{(a-r)(a+r)}{(a+r)^2} = \frac{a-r}{a+r}$$
whence the inequality would follow as was written in the solution.
Any help would be greatly appreciated.
 A: One approach might be to observe that 
$$
\varphi(r,\theta) = \frac{a^2-r^2}{2\pi}\int_0^{2\pi} \frac{|M|}{a^2-2ar\cos(\xi - \theta)+r^2 }\mathrm{d}\xi
$$
solves the boundary value problem with the boundary condition of constant value $ | M | $. If you have already proven the uniqueness of solutions of the boundary value problem, then since the constant function $ \varphi(r,\theta) = |M| $ is a solution, you can conclude the value of the integral is $ |M| $.
Otherwise, you can just compute the integral. If you know some tools from complex analysis, then it's not a hard contour integral. Otherwise, I think there is some reasonable $ u $-substitution that works. For example, using the double angle formula for $ \cos $
$$
\int \frac{1}{a + b \cos(\theta)} d \theta = \int \frac{1}{a+b \; (2 \cos^2(\theta/2) - 1)} d \theta
$$
Then substituting $ u  = \tan(\theta/2) $ seems to go somewhere.
Note again that the value of the integral is $ |M| $, which suggests that any approach involving upper-bounding the kernel $ \frac{1}{a^2-2ar\cos(\theta-\xi)+r^2 } $ won't work. 
A: I've played with this for a while to see if it could be done with some basic estimates on the integral. I haven't had any luck and, as user85055 said, given that
$$\frac{a^2-r^2}{2\pi}\int_0^{2\pi} \frac{M}{a^2-2ar\cos(\theta-\xi) + r^2} \ d\xi = M,$$
we won't be able to derive the result simply by obtaining upper bounds on the Poisson kernel. 
Given this, I think that the best way to solve this problem would be to essentially rederive the maximum principle (at least in the special case where the domain is a disk). Fortunately, this is rather straightforward. I'd be happy to sketch the argument, but I imagine you could find it easily. 
Edit: I'm going to sketch the argument which I'm suggesting. First, we'll prove a version of the maximum principle and then a maximum principle.
Pick some disk $D_\rho(p) \subset D_a(0)$. Choose polar coordinates centered at $p$ and use the Poisson kernel representation formula to get
$$u(p) = \frac{1}{2\pi}\int_0^{2\pi} u(\rho,\theta) \ d\theta.$$
That is, the value of $u$ at the center of any disk is equal to its average over the circle bounding the disk. Now we're ready to prove the result.
So, suppose that $u$ exceeds $M$ at some point in the interior. Then, $u$ must attain its max, call it $A>M$, at some point in the interior of the disk, say $q \in D_a(0)$. Then, pick $0<R<\mathrm{dist}(q,\partial D_a(0))$ and notice that
$$A = u(q) = \frac{1}{2\pi}\int_0^{2\pi} u(R,\theta) \ d\theta \leq A.$$
Notice that the inequality holds as the the average of $u$ over the circle can't exceed the maximum of $u$ over the whole disk. This implies that $u\equiv A$ on $\partial D_R(q)$. Repeating this argument and covering the disk with circles implies that $u\equiv A$ on $D_a(0)$. However, this is a problem as it is not consistent with the BCs; in particular, it implies that $u$ is not continuous on $\overline{D_a(0)}$.
