---
title: "Minimum of x²+y²+z² when 1/x+1/y+1/z=1"
subject: "Mathematics"
chapter: 1
key: "semester_2/mathematics:minimum-x2-y2-z2-when"
source: https://www.bctengineer.com/q/semester_2/mathematics/minimum-x2-y2-z2-when
site: BCT Engineering
---

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

Mathematics · Chapter 1: Calculus of Two or more variables

- Asked in: 2069 Poush, 5 marks
- Video: [Maximum and Minimum of  by using Lagrange's Method of Undetermined  Multipliers problem-1](https://www.youtube.com/watch?v=HanujdaUgXQ) — TEAM Education
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/minimum-x2-y2-z2-when
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-1

## Question

Find the minimum value of the function F(x,y,z) = x²+y²+z² when 1/x + 1/y + 1/z = 1.

## Short answer

Apply Lagrange multipliers: ∂L/∂x = 2x−λ/x²=0 etc., giving x³=y³=z³, so x=y=z. The constraint 1/x+1/y+1/z=1 yields x=y=z=3. The minimum value of x²+y²+z² is 27.

## Answer

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).

```text
∂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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For constraint 1/x+1/y+1/z = 1, ∂L/∂x = 0 gives:
   A. 2x³ = −λ
   B. 2x = λ
   C. 2x = λ/x²
   D. 2x = −λ/x
   Answer: C. 2x − λ/x² = 0, so 2x³ = λ. (Key point: Lagrange equations)

2. From 2x³ = 2y³ = 2z³ = λ we get:
   A. x = y = z = 1
   B. x = y = z = 3
   C. x = y = z = 1/3
   D. x = y = z = 9
   Answer: B. Real cube roots equal; 3/x = 1 gives x = 3. (Key point: Equal values)

3. The minimum value is:
   A. 3
   B. 27
   C. 9
   D. 81
   Answer: B. 9+9+9 = 27. (Key point: Minimum value)

4. Why is 27 a minimum?
   A. Because λ > 0
   B. Because the constraint is linear
   C. QM ≥ HM: the harmonic mean is 3, so √((x²+y²+z²)/3) ≥ 3
   D. AM ≥ GM only
   Answer: C. Gives x²+y²+z² ≥ 27, equality at x = y = z = 3 (for positive numbers). (Key point: Why minimum)

## Close variants set in other exams

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