One can start from Doob's martingale inequality, which states that for every submartingale $(Y_n)_{n\geqslant0}$ and every $y\gt0$,
$$
\mathrm P\left(\max\limits_{0\leqslant k\leqslant n}Y_k\geqslant y\right)\leqslant\frac{\mathrm E(Y_n^+)}y\leqslant\frac{\mathrm E(|Y_n|)}y.
$$
Applying this to $Y_n=(X_n+z)^2$ for some $z\gt0$ and to $y=(x+z)^2$ for some $x\gt0$, one gets
$$
\mathrm P\left(\max\limits_{0\leqslant k\leqslant n}X_k\geqslant x\right)\leqslant\mathrm P\left(\max\limits_{0\leqslant k\leqslant n}Y_k\geqslant y\right)\leqslant C_n(z),
$$
with
$$
C_n(z)=\frac{\mathrm E(|Y_n|)}{y}=\frac{\mathrm E(X_n^2)+z^2}{(x+z)^2}.
$$
Finally, for $z=\dfrac{\mathrm E(X_n^2)}{x}$, $C_n(z)=\dfrac{\mathrm E(X_n^2)}{\mathrm E(X_n^2)+x^2}$ hence the proof is complete.