In the first proof, x+y+z= 1, x2+y2+z2=0.5, X2+Y2+( 1-X-Y) 2 = 0.5, which is arranged into a quadratic equation about Y.

2y2-2 (1-x) y+2x2-2x+0.5 = 0, ∫y∈R, so δ≥ 0.

∴ 4 (1-x) 2-4× 2 (2x2-2x+0.5) ≥ 0, and 0≤x≤2\3, ∴ x ∈ [0 0,2 \ 3]

Y, z ∈ [0 0,2 \ 3] can be obtained in the same way.