David sent in this solution, using the hints we gave you.

\sqrt{a} + \sqrt{b}

rational

\Rightarrow (\sqrt{a} + \sqrt{b})^{2} = a + b + 2 \sqrt{a b}

rational

\Rightarrow 2 \sqrt{a b}

rational

\Rightarrow \sqrt{a b}

rational

\Rightarrow a + \sqrt{a b}

rational
i.e.,

\sqrt{a} (\sqrt{a} + \sqrt{b})

rational

\Rightarrow \sqrt{a}

rational (and so

a

is a square)

\Rightarrow \sqrt{b}

is also rational and hence

b

is a square.

For the second part:

\sqrt{a} + \sqrt{b} + \sqrt{c}

rational

\Rightarrow (\sqrt{a} + \sqrt{b} + \sqrt{c})^{2}

rational

\Rightarrow \sqrt{a b} + \sqrt{b c} + \sqrt{c a}

rational

(\sqrt{a b} + \sqrt{b c} + \sqrt{c a})^{2} = a b + b c + c a + 2 \sqrt{a b c} (\sqrt{a} + \sqrt{b} + \sqrt{c})

so

\sqrt{a b c}

is also rational

\Rightarrow \sqrt{a} (\sqrt{a b} + \sqrt{b c} + \sqrt{c a}) - \sqrt{a b c} = a (\sqrt{b} + \sqrt{c})

is rational
so

\sqrt{b} + \sqrt{c}

is rational and

\sqrt{a}

is rational, and use the above.