Example 1 Let $x, y, z$ be non-negative real numbers, $x+y+z=1$, find the maximum and minimum values of $F=2 x^{2}+y+3 z^{2}$.


First, using substitution and elimination, we have $y=1-x-z$, so
$$\begin{aligned}
F & =2 x^{2}+1-x-z+3 z^{2} \\
& =2\left(x-\frac{1}{4}\right)^{2}+3\left(z-\frac{1}{6}\right)^{2}+\frac{19}{24} \\
& \geqslant \frac{19}{24}
\end{aligned}$$

Also, $F\left(\frac{1}{4}, \frac{7}{12}, \frac{1}{6}\right)=\frac{19}{24}$, so the minimum value of $F$ is $\frac{19}{24}$.
Next, we use the method of successive extrema to find the maximum value of $F$. Fix the variable $z$, then $x+y=1-z$ (a constant).

For $F=2 x^{2}+y+3 z^{2}$, since $z$ is a constant, we only need to find the maximum value of $2 x^{2}+y=A$, where $x+y=1-z$. For convenience, let $1-z=t$, then $x+y=t, 0 \leqslant x, y \leqslant t \leqslant 1, t$ is a constant.

Since $A=2 x^{2}+y=2 x^{2}+t-x$ (substitution and elimination), noting that $0 \leqslant x \leqslant t \leqslant 1$, and the quadratic function opens upwards, the maximum value is not at the vertex, so $A$ can only take its maximum value at $x=0$ or $x=t$. Therefore,
$$g(z)=A_{\max }=\max \left\{t, 2 t^{2}\right\}=\left\{\begin{array}{ll}
t & \left(0 \leqslant t \leqslant \frac{1}{2}\right) \\
2 t^{2} & \left(\frac{1}{2} \leqslant t \leqslant 1\right)
\end{array}\right.$$

Restoring to the original variables, we have
$$g(z)=A_{\max }=\max \left\{1-z, 2(1-z)^{2}\right\}=\left\{\begin{array}{ll}
1-z & \left(\frac{1}{2} \leqslant z \leqslant 1\right), \\
2(1-z)^{2} & \left(0 \leqslant z \leqslant \frac{1}{2}\right)
\end{array}\right.$$

Now, find the maximum value of $g(z)$.
When $0 \leqslant z \leqslant \frac{1}{2}$, $F \leqslant g(z)+3 z^{2}=2(1-z)^{2}+3 z^{2}=5 z^{2}-4 z+2 \leqslant 2$.
When $\frac{1}{2} \leqslant z \leqslant 1$, $F \leqslant g(z)+3 z^{2}=(1-z)+3 z^{2} \leqslant 3$.
Therefore, for all $x, y, z$, we always have $F \leqslant 3$, with equality when $x=y=0, z=1$. So the maximum value of $F$ is 3.

In summary, the minimum value of $F$ is $\frac{19}{24}$, and the maximum value is 3.