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.
 A: 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.
A: For $\sqrt{a}+\sqrt{b}+\sqrt{c}$ to be rational, $ab$ being a perfect square is only a necessary condition, but it is not sufficient (your proof is correct, only the conclusion that it is sufficient is wrong). The sufficient and necessary condition is $a,b,c$ are all perfect squares (easy to see that it is sufficient and I'm going to prove it is necessary). 
Symmetry lets us conclude $ab=A^2,bc=B^2,ca=C^2$.
Let $(a,b)=d$. Then $a=da_1, b=db_1$ and since $d^2a_1b_1=A^2$, we have $a=da_2^2, b=db_2^2$.  
$bc=B^2\implies db_2^2c=B^2\implies dc=b_3^2\tag{1}$  
Let $(c,d)=D$. Then $c=Dc', d=Dd'$. $(1)\implies c=Dc_1^2, d=Dd_1^2$. 
So $(a,b,c)=(D(d_1a_2)^2,D(d_1b_2)^2,Dc_1^2)$  
$$\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{D}(|d_1a_2|+|d_1b_2|+|c_1|)\in\mathbb Q\iff \sqrt{D}\in\mathbb Q$$$$\iff \sqrt{D}\in\mathbb Z\iff D=D'^2$$  
I used the fact that $\sqrt{a},a\in\mathbb Z$ is either an integer or irrational. If it is rational but not an integer, then $\sqrt{a}=\frac{a'}{b'}, a'\nmid b'\implies a=\frac{a'^2}{b'^2}$.
Contradiction, since LHS is an integer and RHS is not ($a'\nmid b'\iff a'^2\nmid b'^2$).
$(a,b,c)=((D'd_1a_2)^2,(D'd_1b_2)^2,(D'c_1)^2)\ \ \ \square$
