Ian H
Lv 7
Ian H asked in Science & MathematicsMathematics · 3 years ago

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)?

Update:

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

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  • 3 years ago
    Favourite 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)

  • JOHN
    Lv 7
    3 years ago

    Answer

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  • atsuo
    Lv 6
    3 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 .

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