<s>[INST] $\mathbb{Q}\left ( \sqrt{2},\sqrt{3},\sqrt{5} \right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3}+\sqrt{5} \right )$ Prove that
$$\mathbb{Q}\left ( \sqrt{2},\sqrt{3},\sqrt{5} \right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3}+\sqrt{5} \right )$$
I proved for two elements, ex, $\mathbb{Q}\left ( \sqrt{2},\sqrt{3}\right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3} \right )$, but I can't do it by the similar method.
 [/INST] This is a boring but purely algebraic derivation.
For any three distinct positive integers $a,b,c$, let $x = \sqrt{a}+\sqrt{b}+\sqrt{c}$, we have
$$\begin{align}
 & (x-\sqrt{a})^2 = (\sqrt{b}+\sqrt{c})^2 = b + c + 2\sqrt{bc}\\
\implies & (x^2 + a - b - c) - 2\sqrt{a}x = 2\sqrt{bc}\\
\implies & (x^2 + a - b - c)^2 - 4\sqrt{a}x(x^2 + a - b - c) + 4ax^2 = 4bc\\
\end{align}
$$
It is easy to check $x^2 + a - b - c \ne 0$. This leads to
$$\sqrt{a} = \frac{(x^2+a-b-c)^2 + 4(ax^2-bc)}{4x(x^2+a-b-c)} \in \mathbb{Q}(x)$$
By a similar argument, we have $\sqrt{b}, \sqrt{c} \in \mathbb{Q}(x)$ and hence
$$\mathbb{Q}(\sqrt{a}, \sqrt{b}, \sqrt{c}) \subset \mathbb{Q}(x) = \mathbb{Q}(\sqrt{a}+\sqrt{b}+\sqrt{c})$$
Since $\sqrt{a} + \sqrt{b} + \sqrt{c} \in \mathbb{Q}(\sqrt{a},\sqrt{b},\sqrt{c})$, we have
$$\mathbb{Q}(\sqrt{a}+\sqrt{b}+\sqrt{c}) \subset \mathbb{Q}(\sqrt{a}, \sqrt{b}, \sqrt{c})$$
Combine these two result, we get 
$$\mathbb{Q}(\sqrt{a}+\sqrt{b}+\sqrt{c}) = \mathbb{Q}(\sqrt{a}, \sqrt{b}, \sqrt{c})$$
</s>