111 Given acute angles $A, B$ satisfy $\tan (A+B)=2 \tan A$, then the maximum value of $\tan B$ is $\qquad$ .
$11 \frac{\sqrt{2}}{4}$ Hint: Since
$$
\begin{aligned}
\tan B & =\tan [(A+B)-A]=\frac{\tan (A+B)-\tan A}{1+\tan (A+B) \tan A} \\
& =\frac{2 \tan A-\tan A}{1+2 \tan ^{2} A}=\frac{\tan A}{1+2 \tan ^{2} A} \\
& \leqslant \frac{\tan A}{2 \sqrt{2} \tan A}=\frac{\sqrt{2}}{4},
\end{aligned}
$$

equality holds if and only if $\tan A=\frac{\sqrt{2}}{2}$. Therefore, the maximum value of $\tan B$ is $\frac{\sqrt{2}}{4}$.