Proofs of the Relationships Among Some Means

by Shannon Umberger


Prove H < G < h < A < g < r < c, where "a" and "b" are real numbers, but "a" does not equal "b".


Step 1: Prove H < G

Note: My proof only works for a, b > 0 and a, b < 0. If you can suggest another proof for all real numbers, please email me at umbie51698@aol.com.

2ab / (a + b)

?
sqrt(ab)

4a^2b^2 / (a^2 + 2ab + b^2)

?
ab

4a^2b^2

?
a^3b + 2a^2b^2 + ab^3

0

?
a^3b - 2a^2b^2 + ab^3

0

?
ab(a^2 - 2ab + b^2)

0

?
ab(a - b)^2

 0

<
(a - b)^2 (for all a, b)

0

<
ab (for a, b > 0 and a, b < 0)


Step 2: Prove G < h

sqrt(ab) 

?
[a + sqrt(ab) + b]/ 3

3*sqrt(ab)

?
a + sqrt(ab) + b

2*sqrt(ab)

?
a + b

4ab

?
a^2 + 2ab + b^2

0

?
a^2 - 2ab + b^2

0

<
(a - b)^2


Step 3: Prove h < A

[a + sqrt(ab) + b]/ 3 

?
(a + b)/ 2

2a + 2*sqrt(ab) + 2b

?
3a + 3b

2*sqrt(ab)

?
a + b

4ab

?
a^2 + 2ab + b^2

0

?
a^2 - 2ab + b^2

0

<
(a - b)^2


Step 4: Prove A < g

(a + b)/ 2 

?
2*(a^2 + ab + b^2) / 3(a + b)

3*(a + b)^2

?
4(a^2 + ab + b^2)

3a^2 + 6ab + 3b^2

?
4a^2 + 4ab + 4b^2

0

?
a^2 - 2ab + b^2

0

<
(a - b)^2


Step 5: Prove g < r

Note: My proof only works for a, b > 0 and a, b < 0. If you can suggest another proof for all real numbers, please email me at umbie51698@aol.com.

2(a^2 + ab + b^2) / 3(a + b)

?
sqrt[(a^2 + b^2) / 2]

4(a^2 + ab + b^2)^2 / 9(a + b)^2

?
(a^2 + b^2) / 2

8(a^2 + ab + b^2)^2

?
9(a^2 + b^2)(a + b)^2

8(a^4 + 2a^3b + 3a^2b^2 + 2ab^3 + b^4)

?
9(a^2 + b^2)(a^2 + 2ab + b^2)

8a^4 + 16a^3b + 24a^2b^2 + 2ab^3 + b^4

?
9a^4 + 18a^3b + 18a^2b^2 + 18ab^3 + 9b^4

0

?
a^4 + 2a^3b - 6a^2b^2 + 2ab^3 + b^4

 0

?
(a^2 - b^2)^2 + 2a^3b - 4a^2b^2 + 2ab^3

0

?
(a^2 - b^2)^2 + 2ab(a^2 - 2ab + b^2)

 0

?
(a^2 - b^2)^2 + 2ab(a - b)^2

0

<
(a^2 - b^2)^2 (for all a, b)

0

<
(a - b)^2 (for all a, b)

0

<
2ab (for a, b > 0 and a, b < 0)


Step 6: Prove r < c

sqrt[(a^2 + b^2) / 2] 

?
(a^2 + b^2) / a + b

(a^2 + b^2) / 2

?
(a^2 + b^2)^2 / (a + b)^2

(a^2 + b^2)*(a + b)^2

?
2*(a^2 + b^2)^2

(a + b)^2

?
2*(a^2 + b^2)

a^2 + 2ab + b^2

?
2a^2 + 2b^2

0

?
a^2 - 2ab + b^2

0

<
(a - b)^2


Therefore, by transitivity, H < G < h < A < g < r < c.


Return to Essay # 3 - Some "Mean" Trapezoids