Show that $\mathbb{E}[\sup_{0 \leq s \leq t} M^4_s] \leq \frac43 \mathbb{E}[M_t \sup_{0 \leq s \leq t} M^3_s] $, where $M = (M_t)_{t \geq 0}$ is a non-negative, continuous sub-martingale starting from $0$.
best I could do was showing the following :
$$\mathbb{E}^2(\sup_{0\leq s \leq t} M_s^2) \leq 4 \mathbb{E}^2(M_t\sup_{0\leq s \leq t} M_s) $$
is there a way to take it from here ?
Set $X_t := \sup_{s \leq t} M_s$. By Doob's inequality, we have
$$\mathbb{P}(X_t \geq r) \leq \frac{1}{r} \int_{\{X_t \geq r\}} M_t \, d\mathbb{P}.$$
Using that
$$\mathbb{E}(X_t^4) = \int_0^{\infty} 4 r^{3} \mathbb{P}(X_t \geq r) \, dr$$
we get
\begin{align*} \mathbb{E}(X_t^4) &\leq 4 \int_0^{\infty} r^2 \int_{\{X_t \geq r\}} M_t \, d\mathbb{P} \, dr \\ &= 4 \int_{\Omega} \int_0^{\infty} r^{2} 1_{[0,X_t]}(r) M_t \, dr \, d\mathbb{P}\\ &= \frac{4}{3} \int_{\Omega} X_t^3 M_t \, d\mathbb{P} \\ &= \frac{4}{3} \mathbb{E}(X_t^3 M_t). \end{align*}