Find the minimum of $(u-v)^2+ (\sqrt{2-u^2}-\frac{9}{v})^2$ for $00$. 
Find the minimum of $\displaystyle (u-v)^2+ \left(\sqrt{2-u^2}-\frac{9}{v}\right)^2$ for
  $0<u<\sqrt{2}$ and $v>0$.

I think I have to use the Arithmetic and Geometric Means Inequalities. Or $\displaystyle \frac{1}{2}(u-v)^2+ \left(\sqrt{2-u^2}-\frac{9}{v}\right)^2 \geq  \frac{1}{4}\left(u-v+\sqrt{2-u^2}-\frac{9}{v}\right)^2$ by Chebychev inequality. Is anyone is able to give me a hint how to finish it?
I think I have to use the Arithmetic and Geometric Means Inequalities.

You can simply think $(u-v)^2+ (\sqrt{2-u^2}-\frac{9}{v})^2$ is the square of distance between two points $a$ and $b$ in the plane, where
$$a=(u,\sqrt{2-u^2}),b=(v,\frac{9}{v}).$$
Clearly, $a$ satisfies $x^2+y^2=2,(0<x<\sqrt{2})$ and $b$ satisfies $y=\frac{9}{x},(x>0)$. Then from the geometry graph, we easily know the minimum distance between the two curves is the distance between $(1,1)$ and $(3,3)$. So the minimum distance is $\sqrt{8}$, and hence the minimum solution is 8 at $u=1,v=3$.
