$\{X_n,n\ge 0\}$ is a martingale and for a $\alpha>1,\mathbb{E}[\lvert X_n\rvert^\alpha]<\infty, \forall n\ge 0$. Let $X_n^\ast=\max\limits_{0\le k\le n}\lvert X_k\rvert$. Prove that
$$\mathbb{E}[X_n^\ast]\le \dfrac{\alpha}{\alpha-1}\mathbb{E}[\lvert X_n\rvert^\alpha]^{1/\alpha}$$
Hint: $\mathbb{E}[X_n^\ast]=\int_0^\infty P[X_n^\ast>t)\,\mathrm{d}t$ and use maximal inequality for submartingale $\lvert X_n\rvert^\alpha$.
My solution: I use the maximal inequality in $L^p$, which shows that $\mathbb{E}[(X_n^\ast)^{p}]^{1/p}\le \dfrac{p}{p-1}(\mathbb{E}[X_n^p])^{1/p}$. It is clear that $\mathbb{E}[X_n^\ast]\le \mathbb{E}[(X_n^\ast)^{p}]^{1/p}$ so we obtain the result.
But the proof of maximal inequality in $L^p$ use lots of the tools like Hölder's inequality and Fubini-Tonelli Theorem and some others. I wonder if there is a more simple proof that use the hint?