Minimum of x²+y²+z² with x+y+z=3a

Minimise f = x² + y² + z² subject to g = x + y + z − 3a = 0.

Lagrange conditions

F = x² + y² + z² + λ(x + y + z − 3a)

∂F/∂x = 2x + λ = 0
∂F/∂y = 2y + λ = 0
∂F/∂z = 2z + λ = 0

So x = y = z = −λ/2. From the constraint 3x = 3a, hence

x = y = z = a (and λ = −2a).

Value

f = a² + a² + a² = 3a².

Nature

By the Cauchy–Schwarz (or QM–AM) inequality, (x + y + z)² ≤ 3(x² + y² + z²), so x² + y² + z² ≥ (3a)²/3 = 3a². Equality holds when x = y = z. So the stationary value is the minimum. (Geometrically, it is the squared distance from the origin to the plane x + y + z = 3a.)

Answer: minimum value = 3a², at x = y = z = a.