Proving the existence of minimum distance between two curves

Question

Consider the following subsets of the plane:$$C_1=\Big\{(x,y)~:~x>0~,~y=\frac1x\Big\} $$and$$C_2=\Big\{(x,y)~:~x<0~,~y=-1+\frac1x\Big\}$$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$$d(P_0,Q_0)\leqslant d(P,Q)\quad\forall ~P\in C_1~\text{and}~Q\in C_2.$$

Source: ISI B MATH 2019 UGB

I took the points $(x,1/x)$ in $C_1$ and $(u,1/u-1)$ in $C_2$ and tried using distance formula but couldn't work it out. Please provide hints/solutions that use highschool mathematics only.

Answer 1

Given the square distance $$ d^2=f(x,u)= \frac{(u-x)^2+u x \left[u x (u-x)^2+2 (u-x)+u x\right]}{u^2 x^2} $$ if we make the derivatives with respect to $x$ and $u$ we get \begin{align} \frac{\partial f}{\partial x} &= -2\ \frac{u^2 x^3-u x^4+u x+u-x}{u x^3}, \\ \frac{\partial f}{\partial u} &= +2\ \frac{u^4 x-u^3 x^2+u x+u-x}{u^3 x}, \end{align} set the numerators equal to $0$ \begin{align} & u^2 x^3-u x^4+u x+u-x = 0, \\ & u^4 x-u^3 x^2+u x+u-x = 0, \end{align} if we add and subtract these equations, we get \begin{align} & u x (u-x) \left(u^2+x^2\right)+2 (u x+u-x) = 0, \\ & u x (u-x)^2 (u+x) = 0. \end{align} Given that $u<0<x,$ the only possible solution of second equation is given by $u=-x$, and substituting in the first equation we have $$ 2 x \left(2 x^4-x-2\right)=0. $$ Again, because $x>0,$ we have to find zeroes of $$ g(x)=2 x^4-x-2. $$ It is easy to see that this function is negative and decreasing from $x=0$ to $x=1/2,$ then it is increasing, and it is positive in $x=2,$ so by the Intermediate value theorem there should be a unique zero between $1/2$ and $2,$ and its value is $$ x=1.1173490365925787\ldots. $$

Answer 2

Answer to the question that the minimum exists and is unique

$C_1$ and $C_2$ are closed subset of $\mathbb R^2$. $P_1=(1,1)$ belongs to $C_1$ while $P_2=(-1,-2) \in C_2$. Therefore $\inf\limits_{(p_1,p_2) \in C_1 \times C_2} d(p_1,p_2) \le d(P_1,P_2)$.

$\Gamma = \{(X,Y) \in C_1 \times C_2 \mid d(X,Y) \le d(P_1,P_2)\}$ is bounded. Being also closed, it is compact and the distance map which is continuous attains its bound on $\Gamma$. Proving that the minimum exists.

It is unique as $\overline{C_1} = \{(x,y) \mid x> 0, y \ge 1/x\}$, $\overline{C_2} = \{(x,y) \mid x< 0, y \le -1+1/x\}$ are convex.

Answer 3

With the distance formula you get a two variables formula, and you must learn the condition for a critical point of a 2-variable function.

On the other hand, I think there is other way, that uses only high school tools. You can construct the vector joining these two generic points and then force to be orthogonal to the tangent vectos of the curves. This way you obtain two equations with two unknowns, and you are done.

Answer 4

Plug in the definition of $y$ and $v$: $$ d = \sqrt{(x-u)^2+(\frac{1}{x} +1 - \frac{1}{u})^2}= \sqrt{x^2 + u^2 -xu -xu + \frac{1}{x^2} + 1 + \frac{1}{u^2} + \frac{2}{x} - \frac{2}{u} - \frac{2}{xu}}$$

AM GM'ing all the non constant terms (Note -u>0):

$$\frac{ x^2 + u^2 - xu -xu + \frac{1}{x^2} + \frac{1}{u^2} + \frac{2}{x} - \frac{2}{u} - \frac{2}{xu} }{9}\geq (2^3)^{\frac19}$$

Hence,

$$ d \geq \sqrt{ 9(2^3)^{\frac19} + 1}$$

Remark: Note that for equality all terms in am gm equal, hence $x=-u$ as noted by enzotib