Inequality with trigonometric functions Find all values for $a$ such that the following inequality holds:
$$\sin^6x + \cos^6x + a\sin x \cos x \ge 0$$

To be fair, I didn't manage to get anything helpful wiht my calculations. I tried to cube the basic trigonometric identity $\sin^2x + \cos^2x = 1$ in order to get a replacement for $\sin^6x + \cos^6x$. Also I tried to check all four quadrants separately, but again I didn't manage to get something.
 A: It is enough to find the minimum value of 
$$
f_a(x):=\cos^6x+\sin^6x+a\cos x\sin x
$$
We have
\begin{eqnarray}
f_a(x)&=&\cos^6x+\sin^6x+a\cos x\sin x\\
&=&(\cos^2x+\sin^2x)^3-3(\cos^4x\sin^2x+\cos^2x\sin^4x)+a\cos x\sin x\\
&=&1-3\cos^2x\sin^2x(\cos^2x+\sin^2x)+a\cos x\sin x\\
&=&1-3\cos^2\sin^2x+a\cos x\sin x\\
&=&1+\frac{a}{2}\sin(2x)-\frac{3}{4}\sin^2(2x)\\
&=&P_a(\sin(2x)),
\end{eqnarray}
where
$$
P_a(t)=1+\frac{a}{2}t-\frac{3}{4}t^2, t \in [-1,1]
$$
It follows that
\begin{eqnarray}
\min_{x\in \mathbb{R}}f_a(x)&=&\min_{t\in [-1,1]}P_a(t)=\min P_a([-1,1])=\min\{\frac{1-2a}{4},\frac{1+2a}{4}\}\\
&=&\frac14\min\{1-2a,1+2a\}=\frac{1-2a+1+2a-|1-2a-(1+2a)|}{8}\\
&=&\frac{2-4|a|}{8}=\frac{1-2|a|}{4}.
\end{eqnarray}
We have
$$
f_a \ge 0 \iff \min f_a \ge 0 \iff 1-2|a|\ge 0 \iff |a|\le \frac12,
$$
i.e. $a \in [-1/2,1/2]$.
A: $$\sin^6x+\cos^6x=\underbrace{(\sin^2x+\cos^2x)}_1\cdot\underbrace{(\sin^4x-\sin^2x\cos^2x+\cos^4x)}_{(\underbrace{\sin^2x+\cos^2x}_1)^2-3\sin^2x\cos^2x}=1-3t^2$$ $=>1+at-3t^2\geqslant0$. Can you take it from here? :-)
