How could you prove?

Question

How could you prove cube root of (xyz)≤(1/3)(x+y+z)(geometric mean≤arithmetic mean)?

Answer 1

Let x+y+z = c

 

f(x) = 27xyz + λ[c - (x+y+z)], Lagrange multipliers

 

fx = 27yz - λ = 0

 

fy = 27xz - λ = 0

 

fz = 27xy - λ = 0

 

λ/27 = xy = xz = yz 

 

=> x = y = z = c/3

 

Therefore, when x=y=z=c/3, you have maximum f(x) = c^3 = (x+y+z)^3.

 

Otherwise,

 

f(x) < (x+y+z)^3

 

Combining the two results: 27xyz ≤ (x+y+z)^3, which is equivalent to

 

(xyz)^(1/3) ≤ (1/3)(x+y+z)^3

Answer 2

First note that x,y, and z must be non-negative for the inequality to hold in general.

 

Let x=a³, y=b³, and z=c³, then

(xyz)^1/3 ≤ (1/3) (x+y+z)  is the same as

abc ≤ (1/3)(a³+b³+c³) or  a³+b³+c³ ≥ 3abc

 

To show this, start with

(a-b)² ≥ 0, so  a²+b² ≥ 2ab.

Similarly,

b²+c² ≥ 2bc  and c²+a² ≥ 2ca.

 

Add these 3 inequalities and divide by 2:

a²+b²+c² ≥ ab+bc+ca

 

Multiply both sides by (a+b+c), which is nonnegative:

(a²+b²+c²)(a+b+c) ≥ (ab+bc+ca)(a+b+c)

 

Multiply both sides out and cancel like terms from both sides. You're left with

a³+b³+c³ ≥ 3abc, or

(xyz)^1/3 ≤ (1/3) (x+y+z).