Example 9 (11th "Hope Cup" Invitational Competition Question) If any real number $x$ makes $\arctan \sqrt{x^{2}+x+\frac{13}{4}} \geqslant$ $\frac{\pi}{3}-a$ hold, then the range of values for $a$ is . $\qquad$
Solve: Fill in $a \geqslant 0$. Reason: Since $\sqrt{x^{2}+x+\frac{13}{4}}=\sqrt{\left(x+\frac{1}{2}\right)^{2}+3} \geqslant \sqrt{3}$, we have arctan $\sqrt{x^{2}+x+\frac{13}{4}} \geqslant \frac{\pi}{3}$.
To ensure the original inequality always holds, we need $\frac{\pi}{3} \geqslant \frac{\pi}{3}-a$, hence $a \geqslant 0$.