I am trying to understand the following proof to the following problem:
The problem is here:
The proof is here:
The link to both is here: https://nrich.maths.org/336
What I don't understand is the second last line. I don't get how we're able to deduce that $\sqrt a (\sqrt a\sqrt b + \sqrt b\sqrt c + \sqrt c\sqrt a) - \sqrt a\sqrt b\sqrt c = rational$, because we don't know if $\sqrt a$ is rational. I also don't get how we're able to deduce that $a(\sqrt b + \sqrt c)$ is rational, because we don't know if $(\sqrt b + \sqrt c)$ is rational.
Could someone explain this to me? Thanks!
It seems a few important steps were skipped in their second part proof, with this resulting in it being hard to follow. The following shows what those steps might have been. Using the various rational quantities they've determined, we get
$$\sqrt{a} + \sqrt{b} + \sqrt{c} = r_1 \implies \sqrt{a} = r_1 - (\sqrt{b} + \sqrt{c}) \tag{1}\label{eq1A}$$
$$\sqrt{ab} + \sqrt{bc} + \sqrt{ca} = r_2 \tag{2}\label{eq2A}$$
$$\sqrt{abc} = r_3 \tag{3}\label{eq3A}$$
where $r_i$, for $1 \le i \le 3$, are positive rational numbers. From their second last line, we then get
$$\begin{equation}\begin{aligned} \sqrt{a}(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) - \sqrt{abc} & = a(\sqrt{b} + \sqrt{c}) \\ (r_1 - (\sqrt{b} + \sqrt{c}))(r_2) - r_3 & = a(\sqrt{b} + \sqrt{c}) \\ r_1r_2 - (\sqrt{b} + \sqrt{c})r_2 - r_3 & = a(\sqrt{b} + \sqrt{c}) \\ r_1r_2 - r_3 & = (a + r_2)(\sqrt{b} + \sqrt{c}) \end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Since the right side is positive, the left side is a positive rational number. Since $a + r_2$ is also a positive rational number, then $\sqrt{b} + \sqrt{c}$ must be a positive rational number. From the left side of the first line of \eqref{eq4A} (or the right side of \eqref{eq1A}), this means $\sqrt{a}$ is also rational. Alternatively, due to the symmetry, you could repeat basically the same procedure to directly prove $\sqrt{b}$ and $\sqrt{c}$ are each rational, so there is no need to use the first part proof.