Finding the integer solutions an equation $
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.

Using the Cauchy-Schwarz inequality is a very clever move: Letting
$${\bf a}:=(3,2,1), \quad{\bf b}:=\bigl(
\sqrt{x+y},\>\sqrt{8-x},\>\sqrt{6-y}\bigr)\tag{1}$$
we get
$$3\sqrt{x+y}+2\sqrt{8-x}+\sqrt{6-y}={\bf a}\cdot{\bf b}\leq|{\bf a}|\>|{\bf b}|=\sqrt{14}\>\sqrt{14}\ ,$$
with equality iff ${\bf b}=\lambda{\bf a}$ for some $\lambda\geq0$ (note that ${\bf a}$ has all components $>0$, and that the components of ${\bf b}$ are $\geq0$). This implies 
$$x+y=9\lambda^2,\quad 8-x=4\lambda^2,\quad 6-y=\lambda^2\ ,$$
which immediately implies $\lambda=1$, and from $(1)$ we then get $x=4$, $y=5$.
The above argument shows that $(4,5)$ is the only real solution (whether integer or not) of the given equation. Using some algebraic elimination procedure would have lead to repeated squarings of the given equation, and only after a lot of computation one would maybe arrive at the simple outcome you have observed at little cost. 
