In this answer I proved the theorem below about positive sums of square roots in ordered fields.
Theorem $\ \sqrt{c_1}+\cdots+\!\sqrt{c_{n}} \in K\ \Rightarrow \sqrt{c_i}\in K\,$ for all $\, i,\:$ if $\,0 < c_i\in K$ an ordered field.
It is instructive to specialize that proof here. Readers unfamiliar with fields should begin here.
Assume that $\, q = \sqrt 3\! +\!\sqrt 7\!+\!\sqrt{21} \in\Bbb Q.\,$
By the Lemma below: $\ \sqrt{7} + \sqrt{21} = q\!-\!\sqrt 3 \in \Bbb Q(\sqrt 3)\,\Rightarrow\, \sqrt{7}, \sqrt{21} \in \Bbb Q(\sqrt 3),\,$ thus
$$\begin{align} \sqrt{7}\, &= \,a\, +\, b\sqrt 3,\ \ \ \ a,\,b\ \in\Bbb Q\\[.3em] \sqrt{21}\, &=\, a' + b'\sqrt 3,\ \ \ a',b'\in\Bbb Q\end{align}\qquad$$
If $\, b\, < 0\,$ then $\,a = \sqrt 7 - b\sqrt 3 = \sqrt 7 + \sqrt{3b^2}\in\Bbb Q\,\Rightarrow\sqrt 7\in \Bbb Q\,$ by the Lemma.
If $\ b'\! < 0\,$ then the same argument allows us to deduce $\sqrt{21}\in \Bbb Q$
Else $\,b,b'\ge 0\,\Rightarrow\,\color{#c00}{1\!+\!b\!+\!b' > 0}\,$ so, by below, we infer $\, \sqrt 3\,\in \Bbb Q$
$\quad\ \ q = \sqrt 3\! +\!\sqrt 7\!+\!\sqrt{21} = a\!+\!a'+(1\!+\!b\!+\!b')\sqrt 3\,\ $ so $\ \sqrt 3 = \dfrac{q\!-\!a\!-\!a'}{\color{#c00}{1\!+\!b\!+\!b'}}\in\Bbb Q$
Lemma $ $ If $\,0< r,s\in K$ then $\, k = \sqrt r\! +\!\sqrt s\in K\Rightarrow\sqrt r,\sqrt s\in K,\,$ for any subfield $\,K\subset \Bbb R$
Proof $\ $ Note $\ k' = \sqrt{r}\!-\!\sqrt{s}\: = \dfrac{\ \ r\, -\ s}{\sqrt{r}\!+\!\sqrt{s}}\in K\ $ by $\,0 < \sqrt r\! +\! \sqrt s\in K,\,$ by $\, \sqrt r,\sqrt s > 0$
Therefore $\ (k+k')/2 = \sqrt r\in K\ $ and $\ (k-k')/2 = \sqrt s\in K$.
Remark $ $ The above is the inductive step of the general proof specialized to the case $\,P(2)\Rightarrow P(3),\,$ where $\,P(n)\,$ denotes the proposition with a sum of $\,n\,$ square roots. The general induction step works precisely the same way.