It is given three positive real numbers \(x,y,z\) such that: \(\cos x+\cos y+\cos z=0\) and \(\cos 3x+\cos 3y+\cos 3z=0\).

Given \( \cos x+\cos y+\cos z=0 \) and \( \cos3x+\cos3y+\cos3z=0 \), prove that \( \cos 2x \cdot \cos 2y \cdot \cos 2z \leq 0 \).

Solution: Using \( \cos3a=4\cos^3a-3\cos a \), we get \( \cos^3x+\cos^3y+\cos^3z=0 \). Since \( \cos x+\cos y+\cos z=0 \), we have \( \cos^3x+\cos^3y+\cos^3z=3\cos x\cos y\cos z \), hence \( \cos x\cos y\cos z=0 \). Assume \( \cos x=0 \), then \( \cos y=-\cos z \), so \( \cos2x\cos2y\cos2z=(2\cos^2x-1)(2\cos^2y-1)(2\cos^2z-1)=-(2\cos^2z-1)^2\leq0 \). Therefore, \( \cos2x\cos2y\cos2z\leq0 \). Q.E.D.