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.

$\bf{Using\; Cauchy\; Schwartz\; Inequality::}$
$$[1^2+1^2]\cdot \left[(u-v)^2+\left(\sqrt{2-u^2}-\frac{9}{v}\right)^2\right]\geq \left[v+\frac{9}{v}-\left(u+\sqrt{2-u^2}\right)\right]^2$$
Now Using $\bf{A.M\geq G.M}\;,$ We get
$\displaystyle v+\frac{9}{v}\geq 6$ and equality hold when $v=3$ and $\displaystyle \left(u+\sqrt{2-u^2}\right)^2\leq 2(u^2+2-u^2)=4$
So we get $\displaystyle \left(u+\sqrt{2-u^2}\right)\leq 2$ and equality hold when $u=1$
So we get $$2\cdot \left[(u-v)^2+\left(\sqrt{2-u^2}-\frac{9}{v}\right)^2\right]\geq \left[6-2\right]^2 = 16$$
So $$\left[(u-v)^2+\left(\sqrt{2-u^2}-\frac{9}{v}\right)^2\right]\geq 8$$
And equality hold ,When $u=3$ and $v=1$
