Example 9 Given real numbers $a, b, c, d$ satisfy $a+b+c+d=3, a^{2}+2 b^{2}+$ $3 c^{2}+6 d^{2}=5$, try to find the range of values for $a$.

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Given that $b+c+d=3-a, 2 b^{2}+3 c^{2}+6 d^{2}=5-a^{2}$. By the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \left(2 b^{2}+3 c^{2}+6 d^{2}\right) \cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right) \\
= & {\left[(\sqrt{2} b)^{2}+(\sqrt{3} c)^{2}+(\sqrt{6} d)^{2}\right] } \\
& \cdot\left[\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{1}{\sqrt{6}}\right)^{2}\right] \\
\geqslant & (b+c+d)^{2}
\end{aligned}$$

That is, $2 b^{2}+3 c^{2}+6 d^{2} \geqslant(b+c+d)^{2}$.
$\therefore 5-a^{2} \geqslant(3-a)^{2}$, solving this gives $1 \leqslant a \leqslant 2$.
$\therefore$ The range of values for $a$ is $[1,2]$.