Two doubts about proof of Doob's Martingale Inequalities Theorem on Jacod-Protter I quote Jacod-Protter.

Theorem Let $M=(M_n)_{n\geq0}$ be a martingale and define $M_n^*=\sup_{j\leq n}|M_j|$. Let $1 < p < \infty$. There exists a constant $c$ depending only on $p$ such that 
  \begin{equation}
\mathbb{E}\{(M_n^*)^p\} \leq c\mathbb{E}\{|M_n|^p\}
\end{equation}

The proof of the above theorem ends in the following manner:
\begin{equation}
\mathbb{E}\{(M_n^*)^p\}\leq\int_0^{\infty}p2p\lambda^{p-2}\mathbb{E}\{|M_n|\mathbb{1}_{\{|M_n|>\frac{\lambda}{2}\}}\}d\lambda = \mathbb{E}\{\int_0^{2|M_n|}2p\lambda^{p-2}d\lambda|M_n|\}=\frac{2^{p}p}{p-1}\mathbb{E}\{|M_n|^{p}\}
\end{equation}
At this point, the theorem is considered as proven and it is stated that 

we showed in the proof of above theorem that the constant $c\leq \frac{2^p p}{p-1}$. With more work one can show that $c^{\frac{1}{p}}=\frac{p}{p-1}$ 

As to the last line above, I would have two doubts and, as a consequence, two observations:
FIRST:
Considering the way in which the theorem is stated, I think it would be correct to state that the constant $c\geq\frac{2^p p}{p-1}$ (that is, $c$ is at least equal to $\frac{2^p p}{p-1}$ for the above theorem inequality to hold true) and NOT the constant $c\leq\frac{2^p p}{p-1}$ (that is, $c$ is at most equal to $\frac{2^p p}{p-1}$ for the above theorem inequality to hold true)
SECOND:
I would say that $c^{\frac{1}{p}}=\frac{2(p)^{\frac{1}{p}}}{{(p-1)}^{\frac{1}{p}}}=2(\frac{p}{p-1})^{\frac{1}{p}}$ and NOT that $c^{\frac{1}{p}}=\frac{p}{p-1}$
Are my observations correct?
 A: For the first doubt :

we showed in the proof of above theorem that the constant $c\leq \frac{2^p p}{p-1}$.

I think that the meaning of this sentence is basically as follows :
"(We want to find $c$ such that $\mathbb{E}\{(M_n^*)^p\} \leq c\mathbb{E}\{|M_n|^p\}$.) We showed in the proof of above theorem that such $c$ can be found at least in $c\leq \frac{2^p p}{p-1}$."
You are correctly claiming that $\mathbb{E}\{(M_n^*)^p\} \leq c\mathbb{E}\{|M_n|^p\}$ holds for every $c$ satisfying $c\color{red}{\ge} \frac{2^p p}{p-1}$, but I don't think that this is what the sentence is trying to say.

For the second doubt :

With more work one can show that $c^{\frac{1}{p}}=\frac{p}{p-1}$

I think that the meaning of this sentence is basically as follows :
"There is another way to show that $\mathbb{E}\{(M_n^*)^p\} \leq (\frac{p}{p-1})^{p}\mathbb{E}\{|M_n|^p\}$ holds (where $(\frac{p}{p-1})^{p}$ is smaller than $\frac{2^pp}{p-1}$)."
I don't think that the sentence means that $c^{\frac 1p}=\frac{p}{p-1}$ follows from $c=\frac{2^pp}{p-1}$.
