Find the minimum value of $(x-y)^2+(x-y+ \frac{1}{y}-\frac{1}{x})^2$ where $x>0>y$

Question

If $\lambda$ denotes the minimum value of $(x-y)^2+(x-y+ \frac{1}{y}-\frac{1}{x})^2$ where $x>0>y$, then find the value of [3$\lambda$] where [.] denotes the greatest integer function.

My approach to this question was to use AM-GM inequalities as both terms in the expression are $>=0$ due to being squared. But I was unsuccessful in finding the minimum value this way. Upon a bit of further inspection, I feel like it could be the distance between two curves but I'm not sure what the two curves would be so please help me with this.

(I only know basic differentiation so far so solutions involving less calculus would be much appreciated)

Answer 1

Like you have mentioned it could be the distance between two curves. However, here its the distance between any two points on a single curve. Consider the function $f(t) = t - \frac{1}{t}$. So here, they are basically asking the distance between points $(x,f(x))$ and $(y,f(y))$.

Considering $y = x - \frac{1}{x}$,we can rearrange it as $x(x-y)=1$. Thats simply a hyperbola. In the question, its mentioned one point has to be taken on the positive x axis and the other on the negative side. Hence we have to find the shortest distance between two branches of the hyperbola and thats the distance between its two vertices.

Answer 2

Denote $t = -xy$, we have: $$\lambda = \underbrace{(x-y)^2}_{\ge 4x(-y)} \left(1+ \left(1+\frac{1}{xy}\right)^2 \right) \ge 4t\left(2-\frac{2}{t} +\frac{1}{t^2}\right) = 4\left( 2t-2+\frac{1}{t}\right)\ge4\left(-2+2\sqrt{2t\cdot\frac{1}{t} } \right) = 8(\sqrt 2 -1) $$

The equality occurs if and only if $x= -y = \sqrt{t} = \frac {1} {\sqrt 2}$ and $[3\lambda_{\min}] =[9.94...] = 9 $.

Answer 3

Continuation of AAM's answer.

Due to AAM's answer, $\lambda=AA'^2$ where $A$ and $A'$ are vertices of the hyperbola $x^2-xy=1$. After a rotation by $22,5^\circ$ about the origin, the hyberbola becomes $\frac{x^2}{2\sqrt2-2}-\frac{y^2}{2+2\sqrt2}=1$. So, $AA'=2\sqrt{2\sqrt2-2}$ and $$\lfloor 3\lambda\rfloor=\lfloor 3AA'^2\rfloor=\lfloor 24(\sqrt2-1)\rfloor=9.$$

The vertices on the original hyperbola are $A(\frac1{\sqrt[4]2},\frac{1-\sqrt2}{\sqrt[4]2})$ and $A'=-A$ by rotating back.