Extreme value of x²+y²+z² with x+y+z=1

Find the extreme value of f = x² + y² + z² subject to x + y + z = 1.

Lagrange conditions

F = x² + y² + z² + λ(x + y + z − 1)

2x + λ = 0
2y + λ = 0
2z + λ = 0

So x = y = z = −λ/2. The constraint gives 3x = 1, so x = y = z = 1/3 (λ = −2/3).

Value

f = 3 × (1/3)² = 3 × 1/9 = 1/3.

Nature

f is the squared distance from the origin to a point of the plane x + y + z = 1. This distance has a minimum (foot of the perpendicular, the point (1/3, 1/3, 1/3), at distance 1/√3) but no maximum. Also, by Cauchy–Schwarz, 1 = (x + y + z)² ≤ 3(x² + y² + z²), so f ≥ 1/3.

Answer: minimum value = 1/3, at (1/3, 1/3, 1/3).