<s>[INST] Inequality $\cos x+\cos y+\cos z=0$ and $\cos{3x}+\cos{3y}+\cos{3z}=0$ prove that $\cos{2x}\cdot \cos{2y}\cdot \cos{2z}\le 0$ Let $x,y,z$ be real numbers such that $\cos x+\cos y+\cos z=0$ and $\cos{3x}+\cos{3y}+\cos{3z}=0$ prove that $\cos{2x}\cdot \cos{2y}\cdot \cos{2z}\le 0$.
 [/INST] Set
$$u:=\cos x , v:=\cos y , w:=\cos z$$
We have 
$$u + v + w = 0$$
and
$$\cos 3x + \cos 3y + \cos 3z = 4u^3 - 3u + 4v^3 - 3v + 4w^3 - 3w = 4u^3 + 4v^3 + 4w^3 = 0$$
So
$$u^3 + v^3 + w^3 = 0$$
Now consider
$$\cos 2x * \cos 2y * \cos 2z = \left(2u^2-1\right)\left(2v^2-1\right)\left(2w^2-1\right)$$
Further, we have
$$\left(u+v+w\right)^3 - u^3 - v^3 - w^3 = 3\left(u^2v+u^2w+v^2u+v^2w+w^2u+w^2v+2uvw\right) = 0$$
$$u^2v+u^2w+u^3+v^2u+v^2w+v^3+w^2u+w^2v+w^3+2uvw = 0$$
$$2uvw = 0$$
WLOG u = 0 ,  v=-w
As $2v^2-1=2w^2-1$ and $2u^2-1 < 0$, the proof is completed.
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