Let $f$ be holomorphic in $\mathbb{D}$ and $0<r<1$. Prove that for every $p\in(0,\infty)$:$$\frac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^pd\theta \geq|f(0)|^p$$
My attempt: First I assume that $p=1$. Then this result is a direct consequence of Cauchy's Integral formula. Now I let $p>0$ and $f$ that doesn't vanish on $\mathbb{D}$. In this case, $g(z)=f^p(z)$ is holomorphic in $\mathbb{D}$ and the result again follows from Cauchy's integral formula.
Now to the general case - let $p>0$ and $f$ holomorphic in $\mathbb{D}$. If $f(0)=0$ then the result is trivial, meaning we can assume $f(0)\neq0$. In this case, since $f$ is holomorphic we have some $\delta>0$ s.t $f(z)\neq 0$ for every $z\in\mathbb{D}_\delta(0)$, and we can use the previous result on $\frac{1}{2\pi}\int_0^{2\pi}|f(\delta e^{i\theta})|^pd\theta$. Now what I want to claim that if we take $\delta<r<1$, we have|: $$\frac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^pd\theta \geq \frac{1}{2\pi}\int_0^{2\pi}|f(\delta e^{i\theta})|^pd\theta \geq|f(0)|^p$$ But I don't know how to show it (or even if it's true).
P.S I know that this has something to do with Hardy's theorem and Volume Integral Means, but I can't use such methods (I'm supposed to use a more fundamental complex analysis tools).
Any hint would be appreciated.
The inequality holds trivially if $f(0) = 0$, so we'll assume that $f(0) \ne 0$. Then $$ \log|f(0)| \le \frac{1}{2 \pi}\int_0^{2 \pi} \log |f(r e^{i \theta}| \, d\theta \, . $$ since $\log |f(z)|$ is subharmonic, or because of Jensen's formula.
It follows that $$ \log\left(|f(0)|^p\right) = p \log|f(0)| \le \frac{p}{2 \pi}\int_0^{2 \pi} \log |f(r e^{i \theta}| \, d\theta \\ = \frac{1}{2 \pi}\int_0^{2 \pi} \log \left(|f(r e^{i \theta}|^p\right) \, d\theta \le \log \left(\frac{1}{2 \pi}\int_0^{2 \pi} |f(r e^{i \theta}|^p \, d\theta \right) \, . $$ The last step is Jensen's inequality applied to the concave function $t \mapsto \log(t)$.