Claim1:All $\sqrt{ab}$,$\sqrt{bc}$ and $\sqrt{ac}$ are integers. Proof:After quadrating both sides we have $a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=2014$ so $\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=x$,$x$ is integer. We have $(\sqrt{ab}+\sqrt{bc})^2=(x-\sqrt{ac})^2$ so we have $ab+bc+2b\cdot \sqrt{ac}=x^2+ac-2x\cdot \sqrt{ac}$ so $\sqrt{ac}$ is rational so it's integer,prove for $\sqrt{bc}$ and $\sqrt{ac}$ are exactly same. Claim2:If $(a,b,c)=1$ we have $a$,$b$,$c$ are perfect squares. Proof:Asumme opposite,let $(a,b)=d$ now using claim1 we have $a=d\cdot p^2$, $b=d\cdot q^2$,if $(a,c)=1$ or $(a,b)=1$ we have contradiction so $(a,c)|p^2$ and because $((a,c),d)=1$ we have that $\frac{p^2}{(a,c)}$ must be perfect square too so $d$ is perfect square so $a$ is perfect square to so as $b$. So if $(a,b,c)=1$ we're done because LHS is integer and RHS is not. If $(a,b,c)\neq 1$ just divide both sides with $\sqrt{(a,b,c)}$ and we have that LHS is integer but RHS will be integer only if $(a,b,c)=2014$ which is apsurd unless $2$ of them are not $0$ so solutions are permutations of$(0,0,2014)$. Edit:As hua1729 pointed out they could be $0$.