Artin (1927) settled Hilbert's 17th problem– any nonnegative polynomial can be written as a sum of squares of rational polynomials. Cassels (1964) proved that a if a polynomial admits an SOS decomposition with $n$ rational polynomials, then it admits one with $n$ polynomials.
That is, Cassels shows that for a field $k,$ if one has $f \in k[x_1,\dots, x_n],$ and $f = \sum_{i=1}^n \left( \frac{g_i}{h_i}\right)^2,$ with $g_i, h_i \in k[x_1,\dots, x_n]$ then $f = \sum_{i=1}^n G_i^2,$ for some $G_i \in k[x_1,\dots, x_n].$ He uses this to prove that $1 + x_1^2 + \dots + x_n^2$ is not a sum of $n$ squares of rational polynomials. I understand Cassel's proof, though for some reason I feel like it's overkill for this example. Hence my question concerns whether one can simplify the proof that $1+x_1^2 + \dots + x_n^2$ cannot be written as a sum of $n$ squares of rational polynomials for the special case where $k=\mathbb{R}.$
To make this even simpler, let's restrict to $\mathbb{R}[x],$ and think about why it's impossible to have $1+x^2 = \frac{g^2(x)}{h^2(x)}$ with $g,h \in \mathbb{R}[x].$ Here's my proposed approach: suppose we make the substitution $x = y/z,$ and then multiply throughout by $z^{\deg(h)}$ to homogenize. Then writing (noting that we must have $\deg(g) = \deg(h)-1$) $$ \overline{h}(y,z) := z^{\deg(h)}h(y/z), \qquad \overline{g}(y,z) := z^{\deg(h)-1}g(y/z), $$ we have $$ (\overline{h}(y,z)y)^2 + (\overline{h}(y,z)z)^2 = \overline{g}(y,z)^2. $$ I feel like there should be a simple reason why this is impossible, but after scouring the internet and staring at the equation for a long time, I can't see why. Any thoughts on where to draw a contradiction on the above would be much appreciated. Two things I've tried are writing the above expression as: $$ (\overline{h}(y,z)y)^2 = (\overline{g}(y,z) - \overline{h}(y,z)z)(\overline{g}(y,z) + \overline{h}(y,z)z)\tag{2} $$ or as $$ 2yz\overline{h}(y,z) = \overline{g}(y,z)^2 - (\overline{h}(y,z)(y+z))^2,\tag{3} $$ but then I am not sure how to proceed!