3. Given positive real numbers $a, b, c$ satisfying $a+b+c=1$, then the maximum value of $a+\sqrt{b}+\sqrt[3]{c}$ is

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3. $\frac{5}{4}+\frac{2 \sqrt{3}}{9}$.

By the AM-GM inequality, we have
$$
\begin{array}{l}
b+\frac{1}{4} \geqslant \sqrt{b} \\
c+\frac{2}{3 \sqrt{3}}=c+\frac{1}{3 \sqrt{3}}+\frac{1}{3 \sqrt{3}} \geqslant 3 \sqrt[3]{c\left(\frac{1}{3 \sqrt{3}}\right)^{2}}=\sqrt[3]{c}
\end{array}
$$

Therefore,
$$
a+\sqrt{b}+\sqrt[3]{c} \leqslant a+b+\frac{1}{4}+c+\frac{2}{3 \sqrt{3}}=\frac{5}{4}+\frac{2 \sqrt{3}}{9},
$$

with equality if and only if $a=\frac{3}{4}-\frac{1}{3 \sqrt{3}}, b=\frac{1}{4}, c=\frac{1}{3 \sqrt{3}}$.
Thus, the maximum value sought is $\frac{5}{4}+\frac{2 \sqrt{3}}{9}$.