Example 3 Given non-negative real numbers $x, y, z, w$ satisfy $x^{2}+y^{2}+z^{2}+w^{2}+x+2 y+3 z+4 w=\frac{17}{2}$, try to find the maximum and minimum values of $x+$ $y+z+w$.

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Example 3 Given non-negative real numbers $x, y, z, w$ satisfy $x^{2}+y^{2}+z^{2}+w^{2}+x+2 y+3 z+4 w=\frac{17}{2}$, try to find the maximum and minimum values of $x+$ $y+z+w$.
The original condition can be transformed into $\left(x+\frac{1}{2}\right)^{2}+(y+1)^{2}+\left(z+\frac{3}{2}\right)^{2}+(w+2)^{2}=16$.
Let $x+\frac{1}{2}=a, y+1=b, z+\frac{3}{2}=c, w+2=d . S=x+y+z+w=a+b+c+d-5$.
By the Cauchy-Schwarz inequality, $(a+b+c+d)^{2} \leqslant 4\left(a^{2}+b^{2}+c^{2}+d^{2}\right)=16 \times 4$.
$a+b+c+d \leqslant 8, x+y+z+w=S \leqslant 3$, and the equality holds if and only if $x=\frac{3}{2}, y=1, z=\frac{1}{2}, w=0$, which means the maximum value of $S=x+y+z+w$ is 3.
Also, $(x+y+z+w)^{2}+4(x+y+z+w) \geqslant x^{2}+y^{2}+z^{2}+w^{2}+x+2 y+3 z+4 w=\frac{17}{2}$, i.e., $S^{2}+4 S \geqslant \frac{17}{2}, S \geqslant-2+\frac{5}{2} \sqrt{2}$, and the equality holds if and only when $x=y=z=0, w=-2+\frac{5}{2} \sqrt{2}$.
In summary, the maximum value of $x+y+z+w$ is 3, and the minimum value is $-2+\frac{5}{2} \sqrt{2}$.