56. Hint: $\frac{a}{a^{2}+1}-\frac{3}{10} \leqslant \frac{18}{25}\left(a-\frac{1}{3}\right) \Leftrightarrow(3 a-1)^{2}(4 a+3) \geqslant 0$.

Generalization: Let real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$, and $a_{1}, a_{2}, \cdots, a_{n} \geqslant -\frac{2 n}{n^{2}-1}(n>1)$, then
$$\sum \frac{a_{1}}{a_{1}^{2}+1} \leqslant \frac{n^{2}}{n^{2}+1}$$

Equality holds if and only if $a_{1}=a_{2}=\cdots=a_{n}=\frac{1}{n}$.
Brief proof: By the method of undetermined coefficients, we can find that
$$\frac{a_{i}}{a_{i}^{2}+1}-\frac{n}{n^{2}+1} \leqslant \lambda\left(a-\frac{1}{n}\right) \quad i=1,2, \cdots, n$$

holds for
$$\lambda=\frac{n^{2}\left(n^{2}-1\right)}{\left(n^{2}+1\right)^{2}}$$

Thus, we can obtain

that is
$$\begin{array}{c} \sum \frac{a_{1}}{a_{1}^{2}+1}-\sum \frac{n}{n^{2}+1} \leqslant 0 \\ \sum \frac{a_{1}}{a_{1}^{2}+1} \leqslant \frac{n^{2}}{n^{2}+1} \end{array}$$