Can I split them into two?

my non-calculus approach: note that $\sqrt{3}+\sqrt{5}=\sqrt{(\sqrt{3}+\sqrt{5})^2}=\sqrt{8+2\sqrt{15}}$ and hence: $\dfrac{\sqrt{3}+\sqrt{5}}{\sqrt{3+\sqrt{5}}}= \sqrt{\dfrac{8+2\sqrt{15}}{3+\sqrt{5}}}$ again, although it is obvious, i tell that $8+2\sqrt{15}>6+2\sqrt{5}\Longrightarrow \sqrt{\dfrac{8+2\sqrt{15}}{3+\sqrt{5}}}>\sqrt{2}$ the given expression can be written as using a similar analysis: $8+2\sqrt{15}<8+2\sqrt{16}=16<20=12+4\sqrt{4}<4(3+\sqrt{5})$ $\dfrac{8+2\sqrt{15}}{3+\sqrt{5}}<4\Longrightarrow \sqrt{\dfrac{8+2\sqrt{15}}{3+\sqrt{5}}}<2$ so $\boxed{\sqrt{2}<\dfrac{\sqrt{3}+\sqrt{5}}{\sqrt{3+\sqrt{5}}}<2}$

First note that $(\sqrt{3} + \sqrt{5})^{2} = 8 + 2\sqrt{15} \gt 6 + 2\sqrt{5} = 2*(3 + \sqrt{5}).$

This implies that $\dfrac{\sqrt{3} + \sqrt{5}}{\sqrt{3 + \sqrt{5}}} \gt \dfrac{\sqrt{2*(3 + \sqrt{5})}}{3 + \sqrt{5}} = \sqrt{2}.$

Next, we look at the function $f(x) = \sqrt{x} + \sqrt{8 - x}$ for $0 \le x \le 8.$ we have that

$f'(x) = \dfrac{1}{2}\left(\dfrac{1}{\sqrt{x}} - \dfrac{1}{\sqrt{8 - x}}\right) = 0$ when $x = 8 - x \Longrightarrow x = 4.$

Since $f(x)$ is continuous on this interval and $f(4) = 4 \gt f(0)$ we know that $f(x)$ is maximized when $x = 4.$

This implies that $(\sqrt{3} + \sqrt{5}) \lt (\sqrt{4} + \sqrt{4}) = 4.$ We also know that $3 + \sqrt{5} \lt 3 + 1 = 4,$ and so

$\dfrac{\sqrt{3} + \sqrt{5}}{\sqrt{3 + \sqrt{5}}} \lt \dfrac{4}{\sqrt{4}} = 2.$

Noting that all other options exclude the interval $(\sqrt{2},2),$ we can conclude that the correct option is

$\boxed{\sqrt{2} \lt \dfrac{\sqrt{3} + \sqrt{5}}{\sqrt{3 + \sqrt{5}}} \lt 2}.$

This problem is really pretty, am I right guys? Cool....refresh