$\mathbb{Q}\left ( \sqrt{2},\sqrt{3},\sqrt{5} \right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3}+\sqrt{5} \right )$ 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.
 A: Notice that $\mathbf{Q}(\sqrt{5})$ is not a subfield of $K=\mathbf{Q}(\sqrt{2},\sqrt{3})$ so $x^{2}-5$ is irreducible in $\mathbf{Q}(\sqrt{2},\sqrt{3})[x]$ and hence $[K(\sqrt{5}):\mathbf{Q}]=8$. This extension is Galois, as the splitting field over $\mathbf{Q}$ of $(x^{2}-5)(x^{2}-3)(x^{2}-2)$. Hence there are 8 distinct automorphisms, sending $\sqrt{5} \to \pm \sqrt{5}$, $\sqrt{3}\to \pm \sqrt{3}$ and $\sqrt{2} \to \pm \sqrt{2}$, to chose an automorphism you pick what it does to $\sqrt{2}$, $\sqrt{5}$ and $\sqrt{3}$. Now $\sqrt{2}+\sqrt{3}+\sqrt{5}$ generates this field if and only if it does not belong to any proper subfield if and only if it's only fixed by the identity automorphism (Galois correspondence).And in fact it is (and it s not hard to check).
A: 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 
$$\mathbb{Q}(\sqrt{a}+\sqrt{b}+\sqrt{c}) = \mathbb{Q}(\sqrt{a}, \sqrt{b}, \sqrt{c})$$
A: It is worth noting from alex.jordan and achille hui's responses that if $\theta = \sqrt{2} + \sqrt{3} + \sqrt{5}$, we explicitly have $$\begin{align*} \sqrt{2} &= \tfrac{5}{3} \theta - \tfrac{7}{72} \theta^3 - \tfrac{7}{144} \theta^5 + \tfrac{1}{576} \theta^7, \\ \sqrt{3} &= \tfrac{15}{4} \theta - \tfrac{61}{24} \theta^3 + \tfrac{37}{96} \theta^5 - \tfrac{1}{96} \theta^7, \\ \sqrt{5} &= -\tfrac{53}{12} \theta + \tfrac{95}{36} \theta^3 - \tfrac{97}{288} \theta^5 + \tfrac{5}{576} \theta^7. \end{align*}$$
