It is given three positive real number `x,y,z` such that : `cos x+cosy+cosz=0` and `cos3x+cos3y+cos3z=0`

 It is given three positive real number `x,y,z` such that : `cos x+cosy+cosz=0` and `cos3x+cos3y+cos3z=0`.

Prove that: `cos 2x.cos 2y.cos 2z`≤ 0

    Solution

Prove that : `cos 2x.cos 2y.cos 2z`≤ 0

Following formula : 

`\cos3a=4\cos^3a-3\cos a` then we can see that:

`\cos3x=4\cos^3x-3\cos x`(1)

`\cos3y=4\cos^3y-3\cos y`(2)

`\cos3z=4\cos^3z-3\cos z`(3)

Make the sum of 3 equations above side to side, we will get:
`4\cos^3x+4\cos^3y+4\cos^3z=0` because `cos x+cosy+cosz=0` and `cos3x+cos3y+cos3z=0`

From `cos x+cosy+cosz=0` then `cos x+cosy=-cosz`

Let's make the 3rd power to the equation above: 
            `(cosx+cosy)^3=-cos^3z`

               `cos^3z+3cosxcosy(cosx+cosy)+cos^3y=-cos^3z`

                `cos^3z-3cosxcosycosz+cos^3y=-cos^3z`

                `cos^3z+cos^3y+cos^3z =3cosxcosycosz`

 But, we knew that: `cos3x+cos3y+cos3z=0`

Then, `3cosxcosycosz=0` ⇒ `cosx=0` or `cosy=0` or `cosz=0`

We assume that: `cosx=0` and `cosy=-cosz`

Then, `cos2xcos2ycos2z=(2cos^2x-1)(2cos^2y-1)(2cos^2z-1)`

            `cos2xcos2ycos2z=-(2cos^2z-1)^2`≤ 0

Hence, the problem is proved.

Solution by: Thin Sokkean