Solve for $a+b$ .

$\begin{aligned}x^2 + y^2 &= 12 \\ (x+y)^2 - 2xy &= 12 \\ 4^2 - 2xy &= 12 \\ 16-2xy &= 12 \\ -2xy &= -4 \\ \therefore xy &= 2 \\ x+y &= 4 \\ \left (\sqrt{x} - \sqrt{y} \right )^2 +2\sqrt{xy} &= 4 \\ \left (\sqrt{x} - \sqrt{y} \right )^2 + 2\sqrt{2} &= 4 \\ \left(\sqrt{x} - \sqrt{y} \right )^2 &= 4-2\sqrt{2} \\ \sqrt{x} - \sqrt{y} &= \pm \sqrt{4-2\sqrt{2}} \\ \therefore \sqrt{x}-\sqrt{y} &= \pm \sqrt{4-\sqrt{8}} \end{aligned}$

Because $x>y$ , then $\sqrt{x} - \sqrt{y} > 0$ . So, $\sqrt{x} - \sqrt{y} = \sqrt{4-\sqrt{8}}$ . Therefore, $a=4$ and $b=8$ $\Longrightarrow a+b = \boxed{12}$ .

Similar solution with @Wildan Bagus Wicaksono 's

$\begin{aligned} x+y & = 4 & \small \color{#3D99F6} \text{Squaring both sides} \\ x^2 + 2xy + y^2 & = 16 & \small \color{#3D99F6} \text{Add } - (x^2+y^2) \text{ on both sides} \\ 2xy & = 16 - \color{#3D99F6} (x^2+y^2) & \small \color{#3D99F6} \text{Given that }x^2 + y^2 = 12 \\ & = 16 - 12 = 4 \\ \implies xy & = 2 \end{aligned}$

Now consider:

$\begin{aligned} z & = \sqrt x + \sqrt y & \small \color{#3D99F6} \text{Squaring both sides} \\ z^2 & = {\color{#3D99F6}x} - 2\sqrt{\color{#D61F06} xy} + \color{#3D99F6} y & \small \color{#3D99F6} \text{Note that }x+y = 4 \\ & = {\color{#3D99F6}4} - 2\sqrt{\color{#D61F06}2} & \small \color{#D61F06} \text{and }xy = 2 \\ \implies z & = \sqrt{4-\sqrt 8} & \small \color{#3D99F6} \text{Taking square root both sides} \end{aligned}$

Therefore, $a+b = 4+8 = \boxed{12}$ .