Show that $ a，b，c, \sqrt{a}+ \sqrt{b}+\sqrt{c} \in\mathbb Q  \implies  \sqrt{a},\sqrt{b},\sqrt{c} \in\mathbb Q $ Assume that $a，b，c, \sqrt{a}+ \sqrt{b}+\sqrt{c} \in\mathbb Q$ are rational，prove $\sqrt{a},\sqrt{b},\sqrt{c}  \in\mathbb Q$,are rational.
I know that can be proved, would like to know that there is no easier way

$\sqrt a + \sqrt b + \sqrt c = p \in \mathbb Q$,
$\sqrt a + \sqrt b  = p- \sqrt c$,
$a+b+2\sqrt a \sqrt b = p^2+c-2p\sqrt c$,
$2\sqrt a\sqrt b=p^2+c-a-b-2p\sqrt c$,
$4ab=(p^2+c-a-b)+4p^2c-4p(p^2+c-a-b)\sqrt c$,
$\sqrt c=\frac{(p^2+c-a-b)+4p^c-4ab}{4p(p^2+c-a-b)}\in\mathbb Q$.

Maybe not easier, but quite elegant :
Suppose that $a,b,c$ are all non zero. Let $K=\mathbb{Q}(\sqrt{a},\sqrt{b},\sqrt{c})$ and $n = [K: \mathbb{Q}]$. Then since $Tr_{K/\mathbb{Q}}(\sqrt{a}) = Tr_{\mathbb{Q}(\sqrt{a})/\mathbb{Q}} \circ Tr_{K/\mathbb{Q}(\sqrt{a})} (\sqrt{a})$, we have
$$ Tr_{K/\mathbb{Q}}(\sqrt{a}) = \begin{cases} 0,& \text{if } \sqrt{a} \notin \mathbb{Q} \\ n\sqrt{a}, &\text{if } \sqrt{a} \in \mathbb{Q}, \end{cases}$$
and same for $\sqrt{b}$ and $\sqrt{c}$.
By hypothesis $\sqrt{a} + \sqrt{b} +\sqrt{c} \in \mathbb{Q}$, so
$$ Tr_{K/\mathbb{Q}}(\sqrt{a}) + Tr_{K/\mathbb{Q}}(\sqrt{b}) + Tr_{K/\mathbb{Q}}(\sqrt{c}) = n\sqrt{a} + n \sqrt{b} + n \sqrt{c}.$$
It is easy to conclude that $\sqrt{a},\sqrt{b},\sqrt{c} \in \mathbb{Q}$.
