Consider the following subsets of the plane: $C_{1}=\{(x,y)|x\gt 0,y=1/x\}$ and $C_{2}=\{(x,y)|x\lt 0,y=-1+1/x\}$ Given two points $P(x,y)$ and $Q(u,v)$ of the plane. Their distance $d(P,Q)$ is defined as: $$d(P,Q)=\sqrt{(x-u)^2+(y-v)^2}$$ Show that there exists unique choice of $P_{0}\in C_{1}$ and $Q_{0}\in C_{2}$ such that $d(P_{0}, Q_{0})\le d(P,Q) \forall P \in C_{1}$ and $Q\in C_{2}$
Attempted:
Essentially all this problem asks for is to prove that there is a lower bound on the distance between the two subsets $C_{1}$ and $C_{2}$ of the plane and that, that minimal distance is achieved at unique $P_{0}$ and $Q_{0}$. To find the minimal distance, we can assert that, it has to be the length of the common normal. From the slopes of the two curves we can say, $P$ must be of the form $(x_{0},1/x_{0})$ and $Q$ of the form $(-x_{0},-1-1/x_{0})$. Now if we plug this into the expression for $d(P_{0},Q_{0})$, we get the following:
$$d(P_{0},Q_{0})=\sqrt{4x^2+1+\frac{4}{x}+\frac{4}{x^2}}$$
To minimise we can set the derivative of $d^2(P_{0},Q_{0})$ with respect to $x$ so get that $2x^4-x-2=0$ which has only one positive solution in $(1,2)$ by the intermediate value theorem, which also proves the uniqueness of the choice of $P_{0}$ and consequently $Q_{0}$. However if I try to apply the AM-GM inequality, and rewrite the expression inside the square root sign as follows, I do not get the condition of equality in which all terms should be equal:
$$x^2+x^2+x^2+x^2+\frac{2}{x^2}+\frac{2}{x^2}+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}\ge 10\cdot2^{1/5}$$
So how to go about it using the AM-GM inequality. Any hints are appreciated. Thanks.
If the two coordinates are $(a,\frac1a)$ and $(-b,-1-\frac1b)$ for $a,b>0$, then the squared distance is $(a+b)^2 + (\frac1a + \frac1b + 1)^2$. We can make the substitution $a = e^u$ and $b = e^v$ to get the function \begin{align} f(u,v) &= (e^u + e^v)^2 + (e^{-u} + e^{-v} + 1)^2 \\ &= 2 e^{-u-v}+2 e^{u+v}+e^{-2 u}+2 e^{-u}+e^{2 u}+e^{-2 v}+2 e^{-v}+e^{2 v}+1 \end{align} which we optimize over all $u,v \in \mathbb R$.
This function is strictly convex, because each term is convex and some are strictly convex, which tells us that the minimizer (if it exists) is unique.
The minimum doesn't have to be achievable in general when optimizing a function over $\mathbb R^2$. However, we can argue that actually, $u,v \in [-2,2]$ (for instance), because $f(0,0)=13$, and one of the terms $e^{2u}, e^{2v}, e^{-2u}, e^{-2v}$ will exceed $13$ if $|u| > 2$ or $|v|>2$. So we can reduce the problem to minimizing $f$ over $[-2,2]^2$, and then a minimizer has to exist by the extreme value theorem.