One-sided martingale maximal inequality ($\mathbb{P}({\max }_{k\le n}{M}_k\ge x)\le V/(V+x^2)$)

For a square-integrable martingale with $M_0=0$ and $V=\mathbb{E}{M}_n^2$, apply the Doob maximal inequality to $(M_k+c{)}^2$ for $c\ge 0$. An upper crossing of $x$ implies a square crossing of $(x+c{)}^2$, giving $(V+c^2)/(x+c{)}^2$. Minimization at $c=V/x$ gives the displayed bound. A one-step martingale taking $x$ with probability $V/(V+x^2)$ and $-V/x$ otherwise attains equality when $V>0$.