# Proofs of the Relationships Among Some Means

## 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