19. Let the three sides of $\triangle ABC$ be $a, b, c$, and $a+b+c=3$. Find the minimum value of $f(a, b, c)=a^{2}+$ $b^{2}+c^{2}+\frac{4}{3} a b c$. (2007 Northern Mathematical Olympiad Problem)
19. Clearly, $a, b, c \in \left(0, \frac{3}{2}\right)$.

Suppose $(a, b, c)$ satisfies the given conditions, then $\left(\frac{a+b}{2}, \frac{a+b}{2}, c\right)$ also satisfies the given conditions. We have
$$\begin{array}{l}
f(a, b, c)-f\left(\frac{a+b}{2}, \frac{a+b}{2}, c\right)=a^{2}+b^{2}+c^{2}+\frac{4}{3} a b c- \\
{\left[2\left(\frac{a+b}{2}\right)^{2}+c^{2}+\frac{4}{3}\left(\frac{a+b}{2}\right)^{2} c\right]=} \\
a^{2}+b^{2}-2\left(\frac{a+b}{2}\right)^{2}+\frac{4}{3} c\left[a b-\left(\frac{a+b}{2}\right)^{2}\right]= \\
\left(\frac{a-b}{2}\right)^{2}-\frac{1}{3} c(a-b)^{2}=(a-b)^{2}\left(\frac{1}{2}-\frac{1}{3} c\right) \geqslant 0
\end{array}$$

Therefore, $f(a, b, c) \geqslant f\left(\frac{a+b}{2}, \frac{a+b}{2}, c\right)$, with equality if and only if $a=b$.
Furthermore,
$$\begin{array}{l}
f\left(\frac{a+b}{2}, \frac{a+b}{2}, c\right)-f(1,1,1)=2\left(\frac{a+b}{2}\right)^{2}+c^{2}+\frac{4}{3}\left(\frac{a+b}{2}\right)^{2} c-\frac{13}{3}= \\
2\left(\frac{3-c}{2}\right)^{2}+c^{2}+\frac{4}{3}\left(\frac{3-c}{2}\right)^{2} c-\frac{13}{3}=\frac{1}{3} c^{3}-\frac{1}{2} c^{2}+\frac{1}{6}= \\
\frac{1}{6}(c-1)^{2}(-2 c+1) \geqslant 0
\end{array}$$

Therefore,
$$f\left(\frac{a+b}{2}, \frac{a+b}{2}, c\right) \geqslant f(1,1,1)=\frac{13}{3}$$

with equality if and only if $c=1$. Hence,
$$f(a, b, c) \geqslant f\left(\frac{a+b}{2}, \frac{a+b}{2}, c\right) \geqslant f(1,1,1)=\frac{13}{3}$$

with equality if and only if $a=b=c=1$.
By the method of local adjustment, we have $[f(a, b, c)]_{\min }=\frac{13}{3}$.