2014/058) Find three positive numbers whose sum is 27 and such that the sum of their squares is as small as possible.

we have (x+y+z)^2 = (x^2+y^2+z^2) + 2(xy+yz+zx)  ... (1)

further 2(x^2+y^2+z^2-xy-xz-yz)= (x-y)^2 + (y-z)^2+(z-x)^2

so 2(xy + yz+xz) = 2(x^2+y^2+z^2) - (x-y)^2 - (y-z)^2 - (z-x)^2

from (1) and above
(x+y+z)^2 = 3((x^2+y^2+z^2) - (x-y)^2 - (y-z)^2 - (z-x)^2
as x+y +z = 27 we have

27+ (x-y)^2 + (y-z)^2 + (x-z)2  = 3(x^2 +y^2 +z^2)

so x^2+y^2 + z^2 is lowest when x = y = z that is x=y=z = 9