Minimum of x²+y²+z² when 1/x+1/y+1/z=1

Minimise F = x² + y² + z² subject to g = 1/x + 1/y + 1/z − 1 = 0. We take x, y, z > 0 (if negative values are allowed, F has no minimum, because the constraint can be met with x, y, z all tiny and of mixed sign).

Lagrange conditions

Let L = x² + y² + z² + λ(1/x + 1/y + 1/z − 1).

∂L/∂x = 2x − λ/x² = 0   →  2x³ = λ
∂L/∂y = 2y − λ/y² = 0   →  2y³ = λ
∂L/∂z = 2z − λ/z² = 0   →  2z³ = λ

So x³ = y³ = z³, hence x = y = z (real values).

Using the constraint

3/x = 1, so x = 3. Thus x = y = z = 3 and λ = 54.

Value

F = 9 + 9 + 9 = 27.

Nature

For positive numbers, the quadratic mean is at least the harmonic mean:

√((x² + y² + z²)/3) ≥ 3/(1/x + 1/y + 1/z) = 3/1 = 3.

Hence x² + y² + z² ≥ 27, with equality at x = y = z = 3. So the value is a minimum.

Answer: minimum value of F = 27, at x = y = z = 3.