Function

$f(x)=\sqrt{3 + \sqrt{5 + \sqrt{x}}}$

$x\ge0$

Smallest value of the function would be $\sqrt{3 + \sqrt{5}}$

Maximum value is infinity.

$a + b\sqrt{c} = \sqrt{3 + \sqrt{5}}$

$a + b\sqrt{5} = \sqrt{3 + \sqrt{5}}$

(a + bsqrt(5))^2 = 3 + sqrt(5)

a^2 + 5b^2 + 2absqrt(5) = 3 + sqrt(5)

a^2 + 5b^2 = 3

sqrt(5) = 2absqrt(5)

2ab = 1

ab = 1/2

You can do it from here... Domain: [a + bsqrt(c), inf)

Get a and b, or just use the sqrt(3 + sqrt(5))