Distance between two curves Consider the following subsets of the plane:
$$
C_1=\{(x,y): x>0, y=\frac{1}{x}\}
$$
and 
$$
C_2=\{(x,y):x<0,y=-1+\frac{1}{x}\}.
$$
Given any two points $P=(x,y)$ and $Q=(u,v)$ of the plane, their distance $d(P,Q)$ is defined by 
$$
d(P,Q)=\sqrt{(x-u)^2+(y-v)^2}.
$$
Show that there exists a unique choice of points $P_0\in C_1$ and $Q_0\in C_2$ such that 
$$
d(P_0,Q_0)\leq d(P,Q) \mbox{ for all } P\in C_1 \mbox{ and } Q\in C_2.
$$
Here if I take two points $p(x,\frac{1}{x}) \in C_1$ and $Q(-x, -1-\frac{1}{x}) \in C_2$ for $x>0$, then I take their distance and take the derivative and prove that there exists a unique point at which the minima occurs.
But I cannot justify my choice of the point $Q \in C_2$, the point could have been $Q(-x^{'} ,-1-\frac{1}{x^{'}})$. But my intuition says that the minimum will occur when I take the same variable for both the points and also it has something to do with the symmetry of the curves. 
My problem is I am not being able to give a mathematical proof of my intuition. 
 A: Let $t= -u>0$, then:
\begin{align}d(P,Q) &=\sqrt{(x-u)^2+\Big({1\over x}-({1\over u}-1)\Big)^2}\\
&= \sqrt{(x+t)^2+\Big(1+{x+t\over xt}\Big)^2}\\
&= \sqrt{a^2+\Big(1+{a\over b}\Big)^2}\\
\end{align}
Now where $a=x+t$ and $b=xy$. Notice that $a^2\geq 4b$ by Am-Gm, so we have:
$$d(P,Q)\geq \sqrt{\underbrace{a^2+\Big(1+{4\over a}\Big)^2}_{f(a)}}$$
So you have to calculate the minumum of $f(a)$ where $a$ is positive number. Now with the derivative of $f$ we see that $a$ satisfies the equation $a^4=16a+4a$ which has exactly one positive solution and thus the conclusion. 
Notice that the task does not ask for explicit $P$ and $Q$. However, no matter what is $a$ we get $b={a^2\over 4}$ so $x={a\over 2}$ and $u=-{a\over 2}$.
A: Your intuition is correct but in order to show it correctly you should use two parameters. 
Let $P=(x,1/x)\in C_1$ and $Q=(t,-1+1/t)\in C_2$ with $x>0$ and $t<0$. Then define 
$$f(x,t):=d(P,Q)^2=(x-t)^2+\left(\frac{1}{x}-\frac{1}{t}+1\right)^2.$$
In order to minimize $f$ we find its critical points, i.e solve
$$\begin{cases}
f_x(x,t)=2(x-t)-2\left(\frac{1}{x}-\frac{1}{t}+1\right)\frac{1}{x^2}=0\\
f_t(x,t)=-2(x-t)+2\left(\frac{1}{x}-\frac{1}{t}+1\right)\frac{1}{t^2}=0
\end{cases}
$$
Note that 
$$0=x^2f_x(x,t)+t^2f_t(x,t)=2(x-t)(x^2-t^2)=2(x-t)^2(x+t)$$
which implies that $t=-x$ (recall that $(x-t)>0$).
Now we proceed with $t=-x$ and we solve $f_x(x,-x)=0$ i.e.
$$P(x):=2x^4-x-2=0$$
which has just one positive root $x_m$ because $P(0)<0$, $P(2)>0$ and $P$ is convex.
Note that $x_m\approx 1.11735$ and the minimum distance is $\approx 3.57459$.  
A: The problem is equivalent to the minimal distance between the two curves
$$y = \frac1x + \frac12, \>\>\>\>\>\>\>   y=\frac1x - \frac12$$
whose minimal distance line passes the origin due to symmetry. Since the distance line is normal to the curves, the following can be established,
$$\frac yx = -\frac1{y'}=x^2\implies y = x^3$$
Plug into the curve $y=\frac1x+\frac12$ to get $x^4-\frac12x - 1=0$, which has the analytic solution 
$$x = \sqrt{\frac a2}\left(1+\sqrt{(2a)^{-3/2}-1}\right)$$
where $a^3+a-\frac1{32}=0$. Considering the smallness of the constant term, we have $a=\frac1{32}$. Then, 
$$x =\frac{1+3\sqrt{7}}8, \>\>\>\>\> y = \left(\frac{1+3\sqrt{7}}8\right)^3$$
which yields the minimal distance 
$$d_{min} = 2\sqrt{x^2+y^2} = 2x\sqrt{2+\frac12x}=\frac{1+3\sqrt{7}}{16}\sqrt{33+3\sqrt7}$$
which is $3.5739$, compared with the exact numerical result $3.5746$.
A: There is a theorem that can be established about the minimum distance of two non-intersecting curves that turns the problem into an algebraic one, provided that one knows how to calculate the derivatives of the curves.
Theorem: For two non-intersecting curves $y_1=f(x)~$,$~y_2=g(x)$, the distance function $$d(x_1,x_2)=\sqrt{(x_1-x_2)^2+(f(x_1)-g(x_2))^2}$$ has critical points at $(x_1,x_2)=(X,Y)$ only if the following two conditions are fulfilled:
1)  The respective tangents of the curves at the points $(X,f(X))~,~(Y,g(Y))$ are parallel.
2)  The line connecting these points is perpendicular to both tangents.
Let's apply this theorem to compute the minimum distance of these two curves. In our case $f(x)=1/x, ~x>0$ and $g(x)=1/x-1~, x<0$. 
To satisfy condition 1) we only need to impose
$$f'(X)=g'(Y)\Rightarrow\frac{1}{X^2}=\frac{1}{Y^2}$$
which implies that $Y=-X$ because of the domain restrictions. 
We find the equation of line connecting the two points to be:
$$y-f(X)=\frac{g(Y)-f(X)}{Y-X}(x-X)$$
Finally to satisfy the second condition we need the slope of this line to be the negative inverse of the slope of the tangent, or more explicitly
$$\frac{f(X)-g(Y)}{X-Y}f'(X)=-1$$
Plugging everything in and imposing $Y=-X$ we obtain
$$\frac{1+\frac{2}{X}}{2X^3}=1\Rightarrow 2X^4-X-2=0$$
The function $u(t)=2t^4-t-2$ has a minimum at $t=1/2$ and also $u(0)=-2<0~~,~~u(1/2)=-\frac{19}{8}<0$ and $\lim_{t\to \pm\infty} u(t)=+\infty$ and therefore it has one positive and one negative root. We see that there is a unique minimum to the distance function since the equation has a unique positive root, at $x\approx 1.11735$. Unfortunately, I do not see an easy way to solve this problem without any knowledge of calculus. Also to prove the theorem stated above, calculus is required.
