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Square of a number

Let a, b, c \in \mathbb{Q} such that \alpha \neq \beta \neq c \neq a. Prove that

    \[\mathcal{A}=\sqrt{\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}}\]

is rational.

Solution

Setting a-b=x ,b-c=y and c-a=z we note that x+y+z=0. Hence,

    \begin{align*} \sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}} &=\sqrt{\frac{x^2y^2+y^2z^2+z^2x^2}{x^2y^2z^2}} \\ &=\sqrt{\frac{(xy+yz+zx)^2-\cancelto{0}{2xyz(x+y+z)}}{x^2y^2z^2}} \\ &=\sqrt{\left(\frac{xy+yz+zx}{xyz}\right)^2} \end{align*}

The result follows.

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