Stop being so mean

From the https://brilliant.org/wiki/power-mean-qagh/ , the quadratic mean is the largest, followed by the arithmetic mean, followed by the geometric mean in a possible sequence of the three means. Also let the distinct numbers be $a$ and $b$ .

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Statement 1:

Assume that we can form an arithmetic sequence such that the three means are consecutive terms and $k$ is the difference between terms so that $\text{Quadratic Mean} = k + \text{Arithmetic Mean}$ $\sqrt{\dfrac{a^2+b^2}{2}} = k + \dfrac{a+b}{2}$ $\dfrac{a^2+b^2}{2} = \left(\dfrac{2k+a+b}{2}\right)^2$ $2(a^2 + b^2) = 4k^2 + 4ka + 4kb + a^2 + b^2 + 2ab$ $a^2 - 2ab + b^2 = 4ka + 4kb + 4k^2$ $\color{#3D99F6}{(a-b)^2 = 4k(a+b+k)}$ Also, $\text{Arithmetic Mean} = k + \text{Geometric Mean}$ $\dfrac{a+b}{2} = k + \sqrt{ab}$ $(a + b - 2k)^2 = 4ab$ $a^2 + 2ab + b^2 - 4ka - 4kb + 4k^2 = 4ab$ $a^2 - 2ab + b^2 = 4ka + 4kb - 4k^2$ $\color{#3D99F6}{(a - b)^2 = 4k(a + b - k)}$

Comparing the two blue lines, we see that it only works if $k=0$ . This means that $a = b$ , which can't happen as they are distinct. Therefore an arithmetic sequence cannot be formed.

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Statement 2:

Assume that we can form a geometric sequence such that the three means are consecutive terms and $k$ is the ratio between terms so that $\text{Quadratic Mean} = k \times \text{Arithmetic Mean}$ $\sqrt{\dfrac{a^2 + b^2}{2}} = \dfrac{k(a + b)}{2}$ $2(a^2 + b^2) = k^2(a+b)^2$ $\color{#20A900}{k^2 = \dfrac{2(a^2+b^2)}{(a+b)^2}}$ Also, $\text{Arithmetic Mean} = k \times \text{Geometric Mean}$ $\dfrac{a+b}{2} = k\sqrt{ab}$ $(a+b)^2 = 4k^2ab$ $\color{#20A900}{k^2 = \dfrac{(a+b)^2}{4ab}}$ Equating the two green statements, and cross-multiplying we get $8ab(a^2 + b^2) = (a + b)^4$ Which simplifies to $(a-b)^4 = 0$ This also implies $a=b$ , meaning that a geometric sequence cannot be formed either