Find the minimum value of $x+2y$ given $\frac{1}{x + 2} + \frac{1}{y + 2} = \frac{1}{3}.$ 
Let $x$ and $y$ be positive real numbers such that
  $$\frac{1}{x + 2} + \frac{1}{y + 2} = \frac{1}{3}.$$Find the minimum value of $x + 2y.$


I think I will need to use the Cauchy-Schwarz Inequality here, but I don't know how I should use it. Can anyone help?
Thanks!

Cauchy-Schwarz implies $$((x+2)+2(y+2))\left(\frac 1{x+2}+\frac 1{y+2}\right)\geq (1+\sqrt{2})^2 $$ $$\Rightarrow x+2y+6\geq 3(1+\sqrt{2})^2,$$where equality is achieved when $$x+2=3(1+\sqrt{2}),y+2=\frac 3{\sqrt{2}}(1+\sqrt{2}).$$ This shows that the minimum of $x+2y$ is $3+6\sqrt{2}.$
