Example 15 (Question from the 13th "Hope Cup" Invitational Competition) Given $ab=1000, a>1, b>1$, what is the maximum value of $\sqrt{1+\lg a}+$ $\sqrt{1+\lg b}$?
Let $\vec{a}=(\sqrt{1+\lg a}, \sqrt{1+\lg b}), \vec{b}=(1,1)$, then
$$
\begin{aligned}
5 & =2+\lg 1000=2+\lg a b=(1+\lg a)+(1+\lg b) \\
& =|\vec{a}|^{2} \geqslant \frac{(\vec{a} \cdot \vec{b})^{2}}{|\vec{b}|^{2}}=\frac{(\sqrt{1+\lg a}+\sqrt{1+\lg b})^{2}}{1^{2}+1^{2}} \\
& =\frac{(\sqrt{1+\lg a}+\sqrt{1+\lg b})^{2}}{2} .
\end{aligned}
$$

Thus, $\sqrt{1+\lg a}+\sqrt{1+\lg b} \leqslant \sqrt{10}$, which means the maximum value sought is $\sqrt{10}$.