Example 6.1.3. Let \( x, y, z \) be positive real numbers with sum 3. Find the minimum of the expression \( x^{2} + y^{2} + z^{3} \).
Solution. Let $a$ and $b$ be positive real numbers. By AM-GM inequality, we have
$$\begin{aligned}
x^{2}+a^{2} & \geq 2 a x \\
y^{2}+a^{2} & \geq 2 a y \\
z^{3}+b^{3}+b^{3} & \geq 3 b^{2} z
\end{aligned}$$

Combining these results yields that $x^{2}+y^{2}+z^{3}+2\left(a^{2}+b^{3}\right) \geq 2 a(x+y)+3 b^{2} z$, with equality for $x=y=a$ and $z=b$. In this case, we could have $2 a+b=x+y+z=3$ (*). Moreover, in order for $2 a(x+y)+3 b^{2} z$ to be represented as $x+y+z$, we must have $2 a=3 b^{2}(\star \star)$. According to $(\star)$ and $(\star \star)$, we easily find out
$$b=\frac{-1+\sqrt{37}}{6}, a=\frac{3-b}{2}=\frac{19-\sqrt{37}}{12}$$
therefore the minimum of $x^{2}+y^{2}+z^{3}$ is $6 a-\left(2 a^{2}+b^{3}\right)$ where $a, b$ are determined as above. The proof is completed.