Let A_1 and A_2 be two arithmetic means, and G_1,G_2,G_3 be three geometric means, of two …

Question

Let $A_1$ and $A_2$ be two arithmetic means, and $G_1,G_2,G_3$ be three geometric means, of two distinct positive numbers $a,b$. Then $G_1^4+G_2^4+G_3^4+G_1^2G_3^2$ is equal to:

Accepted Answer

For $a,A_1,A_2,b$ in A.P.: $A_1+A_2 = a+b$. For $a,G_1,G_2,G_3,b$ in G.P. with ratio $(b/a)^{1/4}$: $G_1=(a^3b)^{1/4}$, $G_2=(a^2b^2)^{1/4}=\sqrt{ab}$, $G_3=(ab^3)^{1/4}$, and $G_1G_3=ab$.

$$G_1^4+G_2^4+G_3^4+G_1^2G_3^2 = a^3b+a^2b^2+ab^3+a^2b^2 = ab(a+b)^2 = G_1G_3(A_1+A_2)^2.$$

Solutions · 1