Example 13.2 Let $a, b, c$ be the three sides of a triangle, satisfying $a+b+c=3$, find the maximum value of the following expression:
$$a^{2}+b^{2}+c^{2}+\frac{4}{3} a b c$$
Let's make the substitution: let $a=x+y, b=y+z, c=z+x$, then $x+y+z=\frac{3}{2}$. After homogenization, the original problem is equivalent to finding the maximum value of $k$ such that the following inequality always holds:
$$\begin{aligned}
f(x, y, z)= & {\left[(x+y)^{2}+(y+z)^{2}+(z+x)^{2}\right] \frac{2(x+y+z)}{3}+} \\
& \frac{4}{3}(x+y)(y+z)(z+x)-k\left[\frac{2(x+y+z)}{3}\right]^{3} \geqslant 0
\end{aligned}$$

Calculating, we get
$$f(1,0,0)=\frac{4}{3}-\frac{8 k}{27}, f(1,1,0)=\frac{32}{3}-\frac{64}{27}, f(1,1,1)=\frac{104}{3}-8 k$$

By the theorem, the necessary and sufficient condition for $f(x, y, z) \geqslant 0$ to always hold is
$$f(1,0,0) \geqslant 0, f(1,1,0) \geqslant 0, f(1,1,1) \geqslant 0$$

Solving these three equations, we get $k \leqslant \frac{13}{2}$, so the maximum value is $\frac{13}{2}$, which is achieved when and only when $a=b=c=1$.