9.34 Find the range of real number $a$ such that the inequality
$$\sin ^{6} x+\cos ^{6} x+2 a \sin x \cos x \geqslant 0$$

holds for all real numbers $x$.
［Solution］ Let $f(x)=\sin ^{6} x+\cos ^{6} x+2 a \sin x \cos x$. Since
$$\begin{aligned}
1 & =\left(\sin ^{2} x+\cos ^{2} x\right)^{3} \\
& =\sin ^{6} x+3 \sin ^{4} x \cos ^{2} x+3 \sin ^{2} x \cos ^{4} x+\cos ^{6} x \\
& =\sin ^{6} x+\cos ^{6} x+3 \sin ^{2} x \cos ^{2} x \\
& =\sin ^{6} x+\cos ^{6} x+\frac{3}{4}+\sin ^{2} 2 x
\end{aligned}$$

Therefore, $f(x)=1-\frac{3}{4} \sin ^{2} 2 x+a \sin 2 x$. 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.