Example 6.1.3 Let $x, y, z>0$, and satisfy $x+y+z=3$, find the minimum value of the following expression
$x^{2}+y^{2}+z^{3} \quad$ (Pham Kim Hung)
Solution: Let $a$ and $b$ be two positive real numbers. By the AM-GM inequality, we have
$$\begin{array}{l}
x^{2}+a^{2} \geq 2 a x \\
y^{2}+a^{2} \geq 2 a y \\
z^{3}+b^{3}+b^{3} \geq 3 b^{2} z
\end{array}$$

Combining these inequalities, we get $x^{2}+y^{2}+z^{3}+2\left(a^{2}+b^{3}\right) \geq 2 a(x+y)+3 b^{2} z$, with equality holding if and only if $x=y=a, z=b$. In this case, we must have $2 a+b=x+y+z=3(*)$. However, to express $2 a(x+y)+3 b^{2} z$ in the form of $x+y+z$, we must have $2 a=3 b^{2}(* *)$. According to (*) and (**), we can easily find
$$b=\frac{-1+\sqrt{37}}{6}, a=\frac{3-b}{2}=\frac{19-\sqrt{37}}{12}$$

Therefore, the minimum value of the expression $x^{2}+y^{2}+z^{3}$ is $6 a-\left(2 a^{2}+b^{3}\right)$, where $a, b$ are determined by the above equations.