<s>[INST] Find conditions on positive integers so that $\sqrt{a}+\sqrt{b}+\sqrt{c}$ is irrational 
Find conditions on positive integers
  $a, b, c$
   so that $\sqrt{a}+\sqrt{b}+\sqrt{c}$ is irrational. 

My solution:
if $ab$ is not the square of an integer,
then the expression is irrational.
I find it interesting
that $c$ does not come into this
at all.
My solution is modeled
(i.e., copied with modifications)
from dexter04's solution
to Prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational
.
Suppose $\sqrt{a}+\sqrt{b}+\sqrt{c} = r$
where $r$ is rational. 
Then, 
$(\sqrt{a}+\sqrt{b})^2 
= (r-\sqrt{c})^2 
\implies a+b+2\sqrt{ab}
 = c+r^2-2r\sqrt{c}$.
So, $a+b-c-r^2+2\sqrt{ab} =-2r\sqrt{c}$.
Let $a+b-c-r^2 = k$, 
which will be a rational number. 
So,
$(k+2\sqrt{ab})^2 = k^2+ 4ab+4k\sqrt{ab} = 4cr^2$
or
$4k\sqrt{ab} = 4cr^2-k^2- 4ab$.
If $ab$ is not a square
of an integer,
then the LHS is irrational
 while the RHS is rational. 
Hence, we have a contradiction.
 [/INST] The answer is that if any of $a, b, c$ is not a square of an integer, then
$\sqrt{a} +\sqrt{b} +\sqrt{c}$ must be irrational.
The proof of the general case is not very easy.  The paper 
Square roots have no unexpected linear relationships by Qiaochu Yuan at 
https://qchu.wordpress.com/2009/07/02/square-roots-have-no-unexpected-linear-relationships/
explains this non-trivial theorem.
</s>