Prove cube of avg of 3 numbers is less than avg of cubes of 3 numbers?

Question

I'm stuck with a journal problem (CMJ), but I can get past it if I can just justify that the cube of the average of three positive real numbers is less than or equal to the average of the cubes of the same three numbers.

i.e. prove ( (x + y + z) / 3) ^ 3 <= (x^3 + y^3 + z^3)/3.

I suspect it is true for any real numbers, but the problem I am working on only requires the numbers to be positive.

I am sure this is either some theorem or some consequence of some theorem. Someone with more stats/number theory experience than me, please help!

Answer 1

Note that; (x+y+z)/3 <= max (x,y,z)

So

( (x + y + z) / 3) ^ 3 <= (max(x,y,z))^3 <= (x^3 + y^3 + z^3)/3

If min(x,y,z) > 0

Since you are only dealing with positives this is fine.

If they are all negative it is simple to see that the reverse inquality must be true.

Answer 2

Straightforward

((x+y+z)/3)^3 = ((x+y+z)^3)/27 so

multiply both sides by 27, divide the right side by 3 and distribute the multiplication

(x+y+z)^3 ≤? 9*x^3 + 9y^3 + 9z^3

Now cube (x+y+z) and get (besides a headache )

x^3 + y^3 + x^3 +3xy^2 + 3yz^2 + 3xz^2 + 3 yx^2 + 3zx^2
+ 3zy^2 + 6xyz

Subtract the x^3, y^3, and z^3 terms from both sides and get

3xy^2 + 3yz^2 + 3xz^2 + 3 yx^2 + 3zx^2 + 3zy^2 + 6xyz
≤? 8x^3 + 8y^3 + 8z^3

Clearly, the *largest* value that the left hand side can attain is when x, y, and z all equal the same value (call it p)

So now we find (substituting p for x, y, and z)

24p^3 ≤? 24p^3 which is clearly true for the equals case.

Now let one of the x, y, or z equal p-k where k is some positive constant k ≤ p.

Plug this in to the big equation (the one stretched over 2 lines above ) do the algebra, and you'll see that the left side is always less than the right side. You can do this with p-k and p-q (where q has the same restriction as k) with 2 of the x, y, z and show the same thing with some straightforward (albeit tedious and messy) algebra.

I'd do it, but I have to go meet some folks about a bit of consulting work.


Doug

Answer 3

{(x+y+z)/3}^3=(x^3+y^3+z^3/27)+{3xy(x+y)+3yz(y+z)+3zx(z+x)}/27>>(x^3+y^3+z^3)/3
if the nos are positive it is evident from the expansion that your statement is true

Answer 4

multiply them out and see what you get...