Prove $\int_{-\pi}^{\pi} \left| 1 - \frac{|z|}{r}e^{iu}\right|^{-q} \mathrm{d}u \leq \left( 1 - \frac{|z|}{r}\right)^{-q-1}$ I have been trying to follow the lines in a proof I saw online. You can read the snippet on page 6. The part of the proof I am stuck on can be stated as follows

Let $r$ ble close to $0$ and $p>1$ then for every $z \in D$ 
  $$
\int_{-\pi}^{\pi}\frac{\mathrm{d}u}{\left| 1 - \frac{|z|}{r}e^{iu}\right|^q}
\leq \left( 1 - \frac{|z|}{r} \right)^{-1-q}
$$

To give some background, the inequality is part of giving an estimate in how the functions in the hardy space are bounded. 

Lemma 1.4 Let $f \in H^p(D)$, Then, for every $z \in D$,
  $$ |f(z)| \leq C_p \frac{\|f\|_{H^p}}{\bigl(1-|z|\bigr)^{1/p}}. $$

The proof starts by defining the Cauchy integral formula, rewriting using $w = re^{it}$ and $z = |z|e^{i\theta}$:
$$
f(z) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{f(re^{it})}{1 - \frac{|z|}{r}e^{i(\theta-t)}}\,\mathrm{d}t
$$
The Author then applies Hölders inequality and the change of variables $u \mapsto \theta - t$:
$$
|f(z)| \leq \|f\|_{H^p} \left( \int_{-\pi}^{\pi}\frac{\mathrm{d}u}{\left|1 - \frac{|z|}{r}e^{iu}\right|^q} \right)^{1/q}
$$
My problem is when the author starts estimating the integral inside the last parenthesis. What it essentially boils down to is showing
$$
\int_{-\pi}^{\pi}\frac{\mathrm{d}u}{\left| 1 - \frac{|z|}{r}e^{iu}\right|^{q}}
\leq \left( 1 - \frac{|z|}{r} \right)^{-1-q}
$$
Does anyone have any idea on how to prove this inequality? In the paper mentioned above, the author proves a similar statement to the one above. However, I was unable to follow that proof as It to me was written somewhat convoluted. I have tried a few things, but alas I have made no progress on the inequality above.
 A: It's regretted, however, that the inequality does not hold.
Take $q=2,|z|/r=1/2$ for instance. Then LHS is $$
\int_{-\pi}^{\pi}\frac{\mathrm{d}u}{\left| 1 - \frac{|z|}{r}e^{iu}\right|^{q}}=
\int_{-\pi}^{\pi}\frac{\mathrm{d}u}{\frac{5}{4}-\cos u}=\frac{8\pi}{3},$$
and RHS is $$
\left( 1 - \frac{|z|}{r} \right)^{-q+1}=2<\frac{8\pi}{3}.
$$
(RHS of your inequality should be $\left( 1 - \frac{|z|}{r} \right)^{-q+1}.$ A typo, I think.)
However we can prove $$
\int_{-\pi}^{\pi}\frac{\mathrm{d}u}{\left| 1 - \frac{|z|}{r}e^{iu}\right|^{q}}\le A\left( 1 - \frac{|z|}{r} \right)^{-q+1},\quad \left(1>\frac{|z|}{r}\ge \frac{1}{4}\right)\tag{1}
$$
where $A$ is a constant independent of $z$, and $(1)$ is enough for us to prove Lemma 1.4. 
Let $\rho=|z|/r>1/4.$ Then 
\begin{align}
|1-\rho e^{iu}|^2&=1+\rho ^2-2\rho \cos u=(1-\rho )^2+4\rho \sin^2 \frac{u}{2}\\
&\ge (1-\rho )^2+\sin^2 \frac{u}{2}\\
&\ge \frac{1}{2}\left(1-\rho +\left|\sin\frac{u}{2}\right|\right)^2,\\
|1-\rho e^{iu}|&=\frac{1}{\sqrt{2}}\left(1-\rho +\left|\sin\frac{u}{2}\right|\right)\\
&\ge \frac{1}{\sqrt{2}}\left(1-\rho +\frac{1}{\pi}|u|\right)\ge \frac{1}{2\pi}\left(1-\rho +|u|\right). 
\end{align}
Therefore we have
\begin{align}
\int_{-\pi}^{\pi}\frac{\mathrm{d}u}{\left| 1 - \rho e^{iu}\right|^{q}}&\le \int_{-\pi}^{\pi}\left(\frac{2\pi}{1-\rho+|u| }\right)^q\mathrm{d}u\\
&=2(2\pi)^q\int_0^{\pi}\frac{1}{(1-\rho+u)^q}\mathrm{d}u\\
&=\frac{2(2\pi)^q}{q-1}\left(\frac{1}{(1-\rho )^{q-1}}-\frac{1}{(\pi+1-\rho )^{q-1}}\right)\\
&\le \frac{A}{(1-\rho )^{q-1}}.
\end{align}
