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.
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. $a \in [-1/2,1/2]$.