II. (25 points) Let the three sides of $\triangle ABC$ be $a, b, c$, and $a+b+c=3$. Find the minimum value of
$$
f(a, b, c)=a^{2}+b^{2}+c^{2}+\frac{4}{3} a b c
$$

(Supplied by Jia Yinghong)
$$
\begin{array}{l}
=\left\{f(a, b, c)=a^{2}+b^{2}+c^{2}+\frac{4}{3} a b c\right. \\
=(a+b+c)^{2}-2(a b+b c+c a)+\frac{4}{3} a b c \\
=9-2\left(a b+b c+c a-\frac{2}{3} a b c\right) .
\end{array}
$$

Since $a, b, c$ are the side lengths of $\triangle ABC$, and $a+b+c=3$, we have $0<a, b, c<\frac{3}{2}$. Therefore,
$$
\begin{array}{l}
\left(\frac{3}{2}-a\right)\left(\frac{3}{2}-b\right)\left(\frac{3}{2}-c\right) \\
\leqslant\left(\frac{\frac{3}{2}-a+\frac{3}{2}-b+\frac{3}{2}-c}{3}\right)^{3}=\frac{1}{8},
\end{array}
$$

which means $a b+b c+c a-\frac{2}{3} a b c \leqslant \frac{7}{3}$.
Thus, $f(a, b, c) \geqslant 9-2 \times \frac{7}{3}=\frac{13}{3}$.
Equality holds if and only if $a=b=c=1$.
Therefore, the minimum value of $f(a, b, c)$ is $\frac{13}{3}$.