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

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.

• Wasn't this question asked some hours ago? – EA304GT Nov 15 '15 at 14:08
• Use the fact that $a^2+b^2\geq 2ab$ for $a,b>0$ – user210387 Nov 15 '15 at 14:28
• the searched minimum is $8$ – Dr. Sonnhard Graubner Nov 15 '15 at 14:52
• I already know this hint by thelink between arithmetic and geometric means. @Rememberme Are you able to explain to me a bit more. – user230283 Nov 15 '15 at 15:21

$\bf{Using\; Cauchy\; Schwartz\; Inequality::}$

$$[1^2+1^2]\cdot \left[(u-v)^2+\left(\sqrt{2-u^2}-\frac{9}{v}\right)^2\right]\geq \left[v+\frac{9}{v}-\left(u+\sqrt{2-u^2}\right)\right]^2$$

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 $v=1$

We want to minimize $$F=-2uv-\frac{18}{v}\sqrt{2-u^2}+v^2+\frac{81}{v^2}+2$$

Then, we have $$\frac{\partial F}{\partial u}=\frac{18u-2v^2\sqrt{2-u^2}}{v\sqrt{2-u^2}}$$

So, if we see $v\gt 0$ as a constant, then we know that $F$ is decreasing for $0\lt u\lt\sqrt{\frac{2v^4}{81+v^4}}$ and is increasing for $\sqrt{\frac{2v^4}{81+v^4}}\lt u\lt \sqrt{2}$. So, $F$ is minimized when $u=\sqrt{\frac{2v^4}{81+v^4}}$. Thus, we can have $$F(u,v)\ge F\left(\sqrt{\frac{2v^4}{81+v^4}},v\right)=\frac{81+v^4}{v^2}-2\sqrt 2\sqrt{\frac{81+v^4}{v^2}}+2$$

By the way, letting $\frac{81+v^4}{v^2}=k$ gives $$(v^2)^2-kv^2+81=0.$$ Considering the discriminant gives that $(-k)^2-4\cdot 81\ge 0\Rightarrow k\ge 18$.

Since $x-2\sqrt 2\sqrt x$ is increasing for $x\gt 2$, we can see that $F$ is minimized when $\frac{81+v^4}{v^2}=18$, i.e. $v=3$.

Thus, the minimum of the expression is $\color{red}{8}$ for $(u,v)=(1,3)$.

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 $u=1,v=3$.

• Could you explain a bit more the fact "we easily know the minimum distance between the two curves is the distance between $(1,1)$ and $(3,3)$" ? Which property allows you to conclude? – user230283 Nov 17 '15 at 1:11
• @J.G Draw a graph, you will get the fact by geometric means. – Chayu Nov 17 '15 at 16:45
• I did both of them, but I don't understand yet. Is it by the fact that the normal vectors coincide, then the distance is minimal at this point? – user230283 Nov 17 '15 at 16:54
• I don't understand the link between this fact and the geometric means? I found normal vector $N_a=(\frac{u}{\sqrt{2-u^2}},1)$ and $N_b=(\frac{9}{v^2},1)$. Could you explain to me a bit more? – user230283 Nov 17 '15 at 16:56
• @J.G Let o denote the origin, then $|oa|=\sqrt{2}, |ob|\leq 3sqrt{2}\Rightarrow|ab|\leq |ob|-|oa|\leq 3\sqrt{2}-\sqrt{2}=\sqrt{8}$. Now, you can derive the equality. – Chayu Nov 17 '15 at 17:15