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 $\cos{2x}\cdot \cos{2y}\cdot \cos{2z}\le 0$.
 A: 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.
A: Setting $\cos x=a$ etc.
We have 
$\displaystyle a+b+c=0\  \ \ \ (1)$
$\displaystyle\implies a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3=(-c)^3+-3ab(-c)+c^3$
$\displaystyle\implies a^3+b^3+c^3=3abc\ \ \ \ (2)$
Again as  $\displaystyle\cos3x=4\cos^3x-3\cos x\implies,$
$\sum\cos3x=0\implies 4(a^3+b^3+c^3)=3(a+b+c)=0\ \ \ \  (3)$
From $\displaystyle (2),(3) 3abc=0$
$\displaystyle\implies $ at least one of $a,b,c$ is zero
If $c=0,$ from $(1), a+b=0\iff b=-a$
$\displaystyle\implies\cos2x\cos2y\cos2z=\prod(2\cos^2x-1)=-(2\cos^2x-1)^2\le0$ as $\cos x=-\cos y\implies \cos^2x=\cos^2y$
