Possible Situation

Let $n$ be any non-perfect square, $n \in \mathbb{N}$ . Then for any $p,q \in \mathbb{N}$ , we have that

$p\sqrt{n} + q\sqrt{n} = (p + q) \sqrt{n}$

which is equivalent to

$\sqrt{p^2 n} + \sqrt{q^2 n} = \sqrt{(p + q)^2 n}$

and clearly none of $p^2 n, q^2 n$ or $(p + q)^2 n$ are zero or perfect squares.

Chose $a=b$ so we have $2\sqrt{a} = \sqrt{c}$ $\sqrt{4a} = \sqrt{c}$ so for all $a=b$ chose $c=4a$ and we will have a solution so long as $a$ is not a perfect square. For example $a=b=2$ , and $c = 8$ produces a valid solution.

Let $a=2$ and $b=8$ . Then $\sqrt 2 + \sqrt 8 = (1+2)\sqrt 2 = \sqrt {18}$ . Yes, it is possible.

Squaring both sides gives $a+2\sqrt{ab}+b = c$ . Since a,b, and c can be any whole numbers, The question can be simplified to
"Are there any whole number values of $a$ and $b$ that aren't zero or perfect squares such that $\sqrt{ab}$ is a whole number"
For $\sqrt{ab}$ to be a whole number $ab$ must be a perfect square, thus the question becomes:
"Are there any whole number values of $a$ , and $b$ that aren't perfect squares such that $ab$ is a perfect square" If the answer is yes then there must be a set (a,b,c) of whole numbers satisfying the equation. Right off the bat you can see $4$ is a perfect square, while 2 isn't, Although $4=2\times2$ . which shows that $a=b=2$ is a solution where $c=8$ , thus proving the statement.