A bound on the maximum of a submartingale I'm trying to prove the following:

If $\{(S_i, \mathcal{F}_i) \mid i \in [N]\}$ is a nonnegative submartingale with $\mathbb{E}S_i^p < \infty$, let $M = \max_i S_i$. Then $\|M\|_p \leq q \|S_N\|_p$.

where $q$ is the conjugate exponent to $p$, that is, $\frac{1}{p} + \frac{1}{q} = 1$. Along the way, I need to prove that if $X, Y$ are nonnegative random variables with finite $p$-th moments and
$$
t \{X > t\} \leq \mathbb{P} Y \{X > t\}
$$
for any $t > 0$ then
$$
\|X\|_p \leq \frac{p}{p-1} \|Y\|_p
$$ (the statement above follows from this via a simple stopping-time argument). I'm somewhat stuck proving this, however. Here's what I've tried so far:
Multiplying through by $pt^{p-2}$ and integrating through in $t$, the left hand side is $\mathbb{E}X^p$, and the right hand side is (after using Tonelli's theorem)
$$
\frac{p}{p-1} \mathbb{E} Y \int_0^{\infty} (p-1)t^{p-2} \{X > t\} dt
$$
Using Holder's inequality gives that this is less than or equal to
$$
\|Y\|_p\frac{p}{p-1} \left[ \mathbb{E} \left(\int_0^{\infty} (p-1)t^{p-2} \{X > t\} dt \right)^q \right]^{1/q}
$$
which is very close to what I want; I want the entire expectation to be $\leq (\mathbb{E}X^p)^{1/q}$ (as ($p-1)q = p$ this is "sort of" right), whence dividing through by a power of $\|X\|_p$ would conclude. Where do I go from here?

As you already figured out, we have by Tonelli's theorem
$$\mathbb{E}(X^p) = \int_0^{\infty} p \cdot r^{p-1} \mathbb{P}(X > r) \, dr.$$
Using the assumption $r \mathbb{P}(X >r) \leq \int_{X>r} Y \, d\mathbb{P}$ this yields
$$\begin{align*} \mathbb{E}(X^p) &\leq \int_0^{\infty} p r^{p-2} \int_{X>r} Y \, d\mathbb{P} \, dr \\ &=p \int \int_0^X r^{p-2} \, dr Y \, d\mathbb{P}\\ &= \frac{p}{p-1} \int X^{p-1} Y \, d\mathbb{P} \end{align*}.$$
Now we apply Hölder's inequality and obtain
$$\mathbb{E}(X^p) \leq \frac{p}{p-1} \|Y\|_q \underbrace{\left(\int X^{(p-1) q} \, d\mathbb{P}\right)^{\frac{1}{q}}}_{\mathbb{E}(X^p)^{1-\frac{1}{p}}}.$$
Dividing by $\mathbb{E}(X^p)^{1-\frac{1}{p}}$ finishes the proof.
