Answer
See proof
Work Step by Step
The tangent line at $(a,b)$ is:
$$y=b+\frac{dy}{dx}_{x=a}(x-a)$$
so:
$$\frac{1}{2\sqrt x}+\frac{\frac{dy}{dx}}{2\sqrt y}=0$$
$$\frac{dy}{dx}=-\frac{\sqrt y}{\sqrt x}=\frac{\sqrt x-\sqrt c}{\sqrt x}$$
so:
$$y=b+\frac{\sqrt a-\sqrt c}{\sqrt a}(x-a)$$
The $x$-intercept of $y$ is:
$$y=0 \to x=\frac{-b\sqrt a}{\sqrt a-\sqrt c}+a$$
The $y$-intercept of $y$ is:
$$x=0 \to y=b-a\frac{\sqrt a-\sqrt c}{\sqrt a}$$
$$x+y=\frac{-b\sqrt a}{\sqrt a-\sqrt c}+a+b-a\frac{\sqrt a-\sqrt c}{\sqrt a}$$ where $$\sqrt a-\sqrt c=-\sqrt b$$
so after simplification:
$$x+y=\sqrt a\sqrt b+a+b+\sqrt a\sqrt b$$
$$x+y=\sqrt a\sqrt b+a+b+\sqrt a\sqrt b=a+2\sqrt a\sqrt b+b=(\sqrt a+\sqrt b)^{2}=(\sqrt c)^{2}=c$$
