If x, y and z are three unit vectors in three-dimensional space, then the minimum value of ||x+y||² + ||y+z||² + ||z+x||² is:

We have ||x+y||² = (x+y).(x+y) = ||x||² + ||y||² + 2x.y = 2 + 2x.y
Similarly, ||y+z||² = 2 + 2y.z and ||z+x||² = 2 + 2z.x
Therefore, ||x+y||² + ||y+z||² + ||z+x||² = 6 + 2(x.y + y.z + z.x)
Let x = (1, 0, 0), y = (1/2, √3/2, 0), z = (1/2, -√3/2, 0)
Then x.y = 1/2, y.z = -1/2, z.x = 1/2
x.y + y.z + z.x = 1/2
||x+y||² + ||y+z||² + ||z+x||² = 6 + 2(1/2) = 7
Consider x.y + y.z + z.x ≤ |x.y| + |y.z| + |z.x|
Since |x.y| ≤ ||x|| ||y|| = 1, similarly |y.z| ≤ 1, |z.x| ≤ 1
Therefore, x.y + y.z + z.x ≤ 3
Minimum value occurs when x.y = y.z = z.x = 1
But this is only possible when x = y = z, which is not possible since x, y, z are unit vectors and they are linearly independent
Consider the case when x, y, z are mutually orthogonal. Then x.y = y.z = z.x = 0
Then ||x+y||² + ||y+z||² + ||z+x||² = 6
Let x = (1,0,0), y=(0,1,0), z=(0,0,1)
Then ||x+y||² + ||y+z||² + ||z+x||² = 2+2+2 = 6
Let θ be the angle between x and y
Then x.y = cosθ
If x, y, z form an equilateral triangle, then x.y = y.z = z.x = -1/2
Then 6 + 2(-3/2) = 3
The minimum value is 3 when x, y, z are equally spaced.