Using various inequalities to prove this martingale inequality? I’ve got another martingale inequality that I would be grateful for a kickstart on.

Suppose $X_t$ is a local martingale such that $|X_t|$ and $\langle X_t\rangle\leq c$ $\forall$ $t\geq 0$ and for some constant
$c\in\mathbb R$. I need to show that $$\mathbb{E}\left(\sup_{t\geq
 0}X_t\right)^4\leq361\mathbb{E}\langle X\rangle_\infty^2.$$

How should I approach this? I asked a question of similar nature previously, so I suspect I will need Cauchy-Schwarz's, Doob's, and Grönwall's inequalities here. However, in the previous question, $X_t$ was given in SDE form, so applying the inequalities made sense — here, all I know is that it is a martingale, so I'm not too sure where to start. Please guide me, thank you!
 A: I am assuming that $X$ is a continuous local martingale such that $X_0=0$ and that $\langle X \rangle$ is the quadratic variation process of $X.$
Applying Ito's formula with $f(x)=x^4$ we have
$$|X_t|^4 = 4 \int_0^t X_t^3 dX_t + 6 \int_0^t X_t^2 d\langle X \rangle_t. \tag{1}$$
Since $X$ is bounded, $E \int_0^t|X_s|^6ds < \infty,$ so the process $\{ \int_0^t X^3_s dX_s, t \geq 0 \}$ is a martingale and therefore, $E \int_0^t X^3_s dX_s = 0.$
Thus, taking expectations in (1)
$$E|X_t|^4 \leq 6 E[\sup_{0 \leq s \leq t} |X_s|^2 \langle X \rangle_t]$$
and by the Holder inequality
$$E|X_t|^4 \leq 6 \left(E\sup_{0 \leq s \leq t} |X_s|^4\right)^{\frac{1}{2}} \left(E\langle X \rangle^2_t \right)^{\frac{1}{2}}. \tag{2}$$
Now (apart from the above) by Doob's martingale inequality
$$E \sup_{0 \leq s \leq t}|X_s|^4 \leq \left(\frac{4}{3}\right)^4 E|X_t|^4. \tag{3}$$
Substituting (2) in (3)
$$E \sup_{0 \leq s \leq t}|X_s|^4 \leq \left(\frac{4}{3}\right)^4 6 \left(E\sup_{0 \leq s \leq t} |X_s|^4\right)^{\frac{1}{2}} \left(E\langle X \rangle^2_t \right)^{\frac{1}{2}}.$$
Thus
$$E \left( \sup_{0 \leq s \leq t}X_s \right)^4 \leq E \sup_{0 \leq s \leq t}|X_s|^4 \leq \left(\frac{4}{3}\right)^8 6^2  E\langle X \rangle^2_t.$$
