I am struggling with this question:
Let $X$ be a continuous local martingale with $X_0=0$, and such that $\mathbb{E} (\langle X \rangle^{p/2}_t) < \infty$, for all $t \geq 0$ and $p \geq 2$. Suppose that $\mathbb{E} (X^2_t) =t$ and $\text{Cov} (X^2_t, \langle X \rangle_t)= 0$ for all $t \geq 0$.
Using the fact that $Y_t = X^4_t - 6X^2_t + 3 \langle X \rangle^2_t$ is a martingale, show that $\mathbb{E} ( X^4_t) \leq 3t^2.$
By the martingale relation and the condition for $X$, I get $$ \mathbb{E}(X^4_t) = -3 \mathbb{E} ( \langle X \rangle^2_t) + 6 \mathbb{E}(X^2_t \langle X \rangle_t)= -3 \mathbb{E} ( \langle X \rangle^2_t) + 6 t\mathbb{E}( \langle X \rangle_t).$$
But how can we bound the moments of $\langle X \rangle$?
Hint: Use $$\mathbb{E}(\langle X \rangle_t)= \mathbb{E}(X_t^2)$$ and $$\mathbb{E}(\langle X \rangle_t^2) \geq (\mathbb{E}(\langle X \rangle_t))^2.$$