<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] 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 $\cos x=-\cos y\implies \cos^2x=\cos^2y$
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