Finding the integer solutions of the equation $3\sqrt {x + y} + 2\sqrt {8 - x} + \sqrt {6 - y} = 14$ $
3\sqrt {x + y}  + 2\sqrt {8 - x}  + \sqrt {6 - y}  = 14
$ . 
I already solved this using the  Cauchy–Schwarz inequality and got $x=4$ and $y=5$. But I'm sure there is a prettier, simpler solution to this and I was wondering if anyone could suggest one.

The Cauchy-Schwarz inequality goes like this:
$LHS \leq \sqrt{3^2+2^2+1^2}\cdot \sqrt{\left(\sqrt{x+y}\right)^2+\left(\sqrt{8-x}\right)^2+\left(\sqrt{6-y}\right)^2}=14=RHS$.
Thus you have equality when $\dfrac{3}{\sqrt{x+y}}=\dfrac{2}{\sqrt{8-x}}=\dfrac{1}{\sqrt{6-y}} \to x=4,y=5$.
