Prove that $\mathbb{Q}[\sqrt{3}, \sqrt{5}] = \mathbb{Q}[\sqrt{3}+ \sqrt{5}]$ Prove that $\mathbb{Q}[\sqrt{3}, \sqrt{5}] = \mathbb{Q}[\sqrt{3}+  \sqrt{5}]$
Take $x \in \mathbb{Q}[\sqrt{3}+  \sqrt{5}]. x = a_x + b_x( \sqrt{3} + \sqrt{5}) = a_x + b_x\sqrt{3} + b_x\sqrt{5} \in \mathbb{Q}[\sqrt{3}, \sqrt{5}] \Rightarrow \mathbb{Q}[\sqrt{3}+  \sqrt{5}] \subset \mathbb{Q}[\sqrt{3}, \sqrt{5}] $
I am having trouble proving the other inclusion. 

While it is possible to show the other inclusion, I think it is far easier to show that the dimensions of each field extension (as vector spaces over $\mathbb{Q}$) are the same (which is $4$ in this case). Then since one vector space is contained within the other, they must be the same if they have the same dimension. Can you handle this from here?
If you wanted to show the other inclusion direction, then you might argue as follows. Note that $(\sqrt 3 + \sqrt 5)^2 = 3 + 2 \sqrt{15} + 5$, and thus $\sqrt{15} \in \mathbb{Q}[\sqrt 3 + \sqrt 5]$. And 
$$\begin{align}
(\sqrt 3 + \sqrt 5 + \sqrt{15})^2 &= 3 + 5 + 15 + 2 \sqrt{15} + 2 \sqrt{45} + 2 \sqrt{75} \\&= 23 + 2 \sqrt{15} + 6 \sqrt{5} + 10 \sqrt{3}.
\end{align}$$
Subtracting $23 + 2 \sqrt{15} + 6(\sqrt{3} + \sqrt{5})$ shows that $4 \sqrt{3} \in \mathbb{Q}[\sqrt{3} + \sqrt{5}]$, and thus $\sqrt{3} \in \mathbb{Q}[\sqrt{3} + \sqrt{5}]$. (And similarly for $\sqrt 5$).
