Define $\overline{X}_{n}:=\max\limits_{1\leq m \leq n} X_{m}$ and let $(X_{n})_{n}$ be a martingale where $X_{0}:=0$ and $EX_{n}^{2}< \infty$
Show that $P(\overline{X}_{n}\geq \lambda) \leq E(X_{n}^{2})/(EX_{n}^{2}+ \lambda^{2})$
As a hint: Consider $(X_{n}+c)^{2}$ and optimize over $c$.
It is clear that $((X_{n}+c)^{2})_{n}$ is a submartingale and when I optimize over $c$, I get:
$X_{n}^{2}+2X_{n}c +c^{2}\Rightarrow 0=2X_{n}+2c=0\Rightarrow -X_{n}=c$
It is also clear I need to use Doob's inequality but I am not sure how to go about this, and how the above optimization helps me.
First observe that for a non-negative $c$, $$ \Pr\left(\bar{X_n}\geqslant\lambda\right)\leqslant \Pr\left(\max_{1\leqslant m\leqslant n}\left(X_m+c\right)^2\geqslant \left(\lambda+c\right)^2\right); $$ apply Doob's inequality to get a control of the latter probability by a term involving $\mathbb E\left[\left(X_n+c\right)^2\right]$. This leads to a bound of $\Pr\left(\bar{X_n}\geqslant\lambda\right)$ valid for every non-negative $c$, namely, $$ \Pr\left(\max_{1\leqslant m\leqslant n}\left(X_m+c\right)^2\geqslant \left(\lambda+c\right)^2\right)\leqslant \frac1{\left(\lambda+c\right)^2}\mathbb E\left[\left(X_n+c\right)^2\right]=:f(c).$$ Then optimize $f(c)$ over $c$.