Modified Doob's $L^1$ inequality

Let $X_n$ be a non-negative submartingale. Show that

for all $\lambda >0$

$$ P(\sup_{k\leq n} X_n \geq 2\lambda) \leq \frac{1}{\lambda} \int_{X_n \geq \lambda} X_n dP$$

In Doob's weak $L^1$ inequality, the right hand side integrates over $\left\lbrace \sup_{k\leq n} X_n \geq 2\lambda \right\rbrace$. I have been stuck on part (1) already. Does anyone have idea how to solve this problem? Thanks!

EDIT: Hint added, consider $Y_k = E(X_n 1_{X_n \geq \lambda} | F_k)$ At present I'm not quite sure how to apply this hint.

A first step could be the following. Let $M_n:=\max_{1\leqslant k\leqslant n}X_k$. Then for any positive real number $t$, $$\tag{*} 2t\mathbb P\left(M_n\gt 2t\right)\leqslant \mathbb E\left[X_n\mathbf 1\left\{M_n>2t\right\}\right]. $$ This is Doob's weak $\mathbb L^1$ inequality.

Now, rewrite the right hand side of (*) as $\int_0^{+\infty}\mathbb P \left(\left\{X_n>x\right\}\cap\left\{M_n>2t\right\}\right)\mathrm dx$. For any real number $a$, this can be bounded by $a \mathbb P \left(\left\{M_n>2t\right\}\right)+\int_a^{+\infty}\mathbb P \left(\left\{X_n>x\right\}\right)\mathrm dx $ which is in turn smaller than $a \mathbb P \left(\left\{M_n>2t\right\}\right)+\mathbb E\left[X_n\mathbf 1\left\{X_n>a\right\}\right]$. We thus got that $$2t\mathbb P\left(M_n\gt 2t\right)\leqslant a \mathbb P \left(\left\{M_n>2t\right\}\right)+\mathbb E\left[X_n\mathbf 1\left\{X_n>a\right\}\right]. $$ Choosing $a=t$ gives the wanted result.