Given that (a ^ 2) + (b ^ 2) + (C ^ 2) = 1, (x ^ 2) + (y ^ 2) + (Z ^ 2) = 1, prove: ax + by + CZ

Given that (a ^ 2) + (b ^ 2) + (C ^ 2) = 1, (x ^ 2) + (y ^ 2) + (Z ^ 2) = 1, prove: ax + by + CZ

Because (a ^ 2) + (b ^ 2) + (C ^ 2) = 1, (x ^ 2) + (y ^ 2) + (Z ^ 2) = 1
(a^2)+(b^2)+(c^2)+(x^2)+(y^2)+(z^2)-2(ax+by+cz)
=(a-x)^2+(b-y)^2+(c-z)^2>=0
Take the known conditions into (a ^ 2) + (b ^ 2) + (C ^ 2) = 1, (x ^ 2) + (y ^ 2) + (Z ^ 2) = 1
2-2(ax+by+cz)>=0
So ax + by + CZ