# Prove (x + y + z)/3 ≥ (xyz)^(1/3) by ALGEBRA like adding 4xy to (x – y)^2 ≥ 0 to get AM–GM for 2 variables to get (x + y)/2 ≥ √(xy)?

Prove (x + y + z)/3 ≥ (xyz)^(1/3)

by ALGEBRA like adding 4xy to

(x – y)^2 ≥ 0 to get AM–GM for 2 variables

(x + y)/2 ≥ √(xy)

### 3 Answers

- husoskiLv 73 years agoFavourite answer
The way I remember has a trick to it. Start by using the 2-variable version to prove AM-GM for n=4:

(x + y + z + w)/4 = (1/2)[(x + y)/2 + (z + w)/2]

(x + y + z + w)/4 ≤ (1/2)[√(xy) + √(zw)] .... apply AM-GM_2 to each term in [] brackets

(x + y + z + w)/4 ≤ √[√(xy) √(zw)] .... apply it once more to remaining terms

(x + y + z + w)/4 ≤ ∜(xyzw)

For three values x,y,z you can invent a fourth value w=∛(xyz), so that:

(x + y + z + w)/4 ≤ ∜(xyzw) .... AM-GM_4, above

(x + y + z + w)/4 ≤ ∜(w³w) = ∜(w⁴) = w

x + y + z + w ≤ 4w

x + y + z ≤ 3w

(x + y + z)/3 ≤ w = ∛(xyz)

- atsuoLv 63 years ago
Assume x,y,z are positive .

Let A = x^(1/3) , B = y^(1/3) and C = z^(1/3) .

So A,B,C are positive .

We know the formula (it stands for any values of A,B,C) :

A^3 + B^3 + C^3 - 3ABC = (A + B + C)(A^2 + B^2 + C^2 - AB - BC - CA)

So

A^3 + B^3 + C^3 - 3ABC = (A + B + C)(A^2 - 2AB + B^2 + B^2 - 2BC + C^2 + C^2 - 2CA + A^2)/2

A^3 + B^3 + C^3 - 3ABC = (A + B + C)[(A - B)^2 + (B - C)^2 + (C - A)^2]/2

(A + B + C) is positive and [(A - B)^2 + (B - C)^2 + (C - A)^2] is non-negative , so

A^3 + B^3 + C^3 - 3ABC is non-negative .

Therefore

A^3 + B^3 + C^3 - 3ABC ≧ 0

A^3 + B^3 + C^3 ≧ 3ABC

x + y + z ≧ 3[x^(1/3)*y^(1/3)*z^(1/3)]

x + y + z ≧ 3(xyz)^(1/3)

(x + y + z)/3 ≧ (xyz)^(1/3)

So it is proved .