= Solution
Put $v=\mathbb E X_n^2$. For $c\geq0$, <conditional Jensen inequality> applied to the <convex function> $x\mapsto(x+c)^2$ shows that
$$
Z_k=(X_k+c)^2
$$
is a nonnegative <submartingale>. Its integrability follows from the square integrability of $X_k$. If $X_k\geq\lambda$, then $Z_k\geq(\lambda+c)^2$, since $c\geq0$. The <Doob maximal inequality for a nonnegative submartingale> yields
$$
\mathbb P\left(\max_{1\leq k\leq n}X_k\geq\lambda\right)
\leq\frac{\mathbb E(X_n+c)^2}{(\lambda+c)^2}
=\frac{v+c^2}{(\lambda+c)^2},
$$
where zero mean removes the cross term. For completeness, the maximal inequality follows by stopping at the first crossing: on the event of a crossing at $k$, the <submartingale> property gives $\mathbb E[Z_n\mathbf1_{\{T=k\}}]\geq\mathbb E[Z_k\mathbf1_{\{T=k\}}]$. Sum over $k\leq n$, and use nonnegativity on the event of no crossing.
The derivative of the last ratio is
$$
\frac{d}{dc}\frac{v+c^2}{(\lambda+c)^2}
=\frac{2(c\lambda-v)}{(\lambda+c)^3}.
$$
For $v>0$, the minimum over $c\geq0$ is attained at $c=v/\lambda$. Substitution gives the <one-sided maximal inequality for a centered square-integrable martingale>:
$$
\boxed{\mathbb P\left(\max_{1\leq k\leq n}X_k\geq\lambda\right)
\leq\frac{v}{\lambda^2+v}.}
$$
If $v=0$, $X_n=0$ <almost surely> and $X_k=\mathbb E(X_n\mid\mathcal F_k)=0$ <almost surely> for every $k\leq n$, so the bound also holds. The optimization is the same one underlying the <Cantelli inequality>, but the <submartingale> argument controls the entire finite-time maximum.
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