$9 \cdot 34$ Find the range of real values of $a$ such that the inequality
$$
\sin ^{6} x+\cos ^{6} x+2 a \sin x \cos x \geqslant 0
$$

holds for all real numbers $x$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
[Solution] Let $f(x)=\sin ^{6} x+\cos ^{6} x+2 a \sin x \cos x$. Since
$$
\begin{aligned}
1 & =\left(\sin ^{2} x+\cos ^{2} x\right)^{3} \\
& =\sin ^{6} x+3 \sin ^{4} x \cos ^{2} x+3 \sin ^{2} x \cos ^{4} x+\cos ^{6} x \\
& =\sin ^{6} x+\cos ^{6} x+3 \sin ^{2} x \cos ^{2} x \\
& =\sin ^{6} x+\cos ^{6} x+\frac{3}{4}+\sin ^{2} 2 x,
\end{aligned}
$$

we have $f(x)=1-\frac{3}{4} \sin ^{2} 2 x+a \sin 2 x$. If $|a| \leqslant \frac{1}{4}$, then
$$
f(x) \geqslant 1-\frac{3}{4} \sin ^{2} 2 x-|a||\sin 2 x| \geqslant 1-\frac{3}{4}-\frac{1}{4}=0
$$

for all real numbers $x$. Conversely, if $|a|>\frac{1}{4}$, take a real number $x_{0}$ such that $a \sin 2 x_{0}=$ $-|a|$, then $\left|\sin 2 x_{0}\right|=1$, and
$$
f\left(x_{0}\right)=1-\frac{3}{4}-|a|<0 .
$$

In summary, $|a| \leqslant \frac{1}{4}$ is the required condition.