Extreme values with two constraints

Find the extreme values of f = x² + y² + z² subject to g₁ = x + y + z − 1 = 0 and g₂ = xyz + 1 = 0.

Lagrange conditions

F = x² + y² + z² + λ(x + y + z − 1) + μ(xyz + 1)

2x + λ + μyz = 0
2y + λ + μxz = 0
2z + λ + μxy = 0

Subtract the first two: 2(x − y) − μz(x − y) = 0, so (x − y)(2 − μz) = 0. Similarly (y − z)(2 − μx) = 0 and (z − x)(2 − μy) = 0.

Case analysis

• If x, y, z were all different, then μz = μx = μy = 2, forcing x = y = z, a contradiction. So at least two are equal.
• Take x = y. Then z = 1 − 2x and x²z = −1:

x²(1 − 2x) = −1
2x³ − x² − 1 = 0
(x − 1)(2x² + x + 1) = 0

2x² + x + 1 has discriminant 1 − 8 < 0 (no real root), so x = 1, giving y = 1, z = −1.

The stationary points are (1, 1, −1), (1, −1, 1), (−1, 1, 1), and at each

x² + y² + z² = 1 + 1 + 1 = 3.

Nature

Since x + y + z = 1, x²+y²+z² = 1 − 2(xy + yz + zx). The numbers x, y, z are roots of t³ − t² + e t + 1 = 0 where e = xy + yz + zx. All three roots are real only if its discriminant −(e + 1)(4e² − 5e + 23) ≥ 0, i.e. e ≤ −1. Hence x² + y² + z² = 1 − 2e ≥ 3, so 3 is the minimum, and there is no maximum (the value can be made as large as we like).

Answer: the extreme value is the minimum 3, attained at (1, 1, −1) and its permutations.