Four points $A$, $B$, $C$ and $D$ are chosen on each of $4$ hyperbola branches of $x^2y^2=1$ (as $y=\dfrac{1}{x}$ and $y=-\dfrac{1}{x}$ combined). Find the minimum area of the quadrilateral.🔑
Solution: Given that all 4 vertices of the quadrilaterial are on the $4$ branches, let $A=(a,\dfrac{1}{a})$, $B=(-b,\dfrac{1}{b})$, $C=(-c,-\dfrac{1}{c})$, and $D=(d,-\dfrac{1}{d})$, where $a,b,c,d>0$.
By the shoelace formula, we have
$$[ABCD]=\dfrac{1}{2}\Big{(}(a\cdot\dfrac{1}{b}-\dfrac{1}{a}\cdot(-b))+((-b)\cdot(-\dfrac{1}{c})-\dfrac{1}{b}\cdot(-c))$$
$$+(-c\cdot(-\dfrac{1}{d})-(-\dfrac{1}{c}\cdot d))+(d\cdot\dfrac{1}{a}-a\cdot(-\dfrac{1}{d}))\Big{)}\tag{1}$$
$$=\dfrac{1}{2}\Big{(}(\dfrac{a}{b}+\dfrac{b}{a})+(\dfrac{b}{c}+\dfrac{c}{b})+(\dfrac{c}{d}+\dfrac{d}{c})+(\dfrac{d}{a}+\dfrac{a}{d})\Big{)}$$
By AM–GM inequality, for $x,y>0$ we have $$\dfrac{x}{y}+\dfrac{y}{x}\ge2\sqrt{\dfrac{x}{y}\cdot\dfrac{y}{x}}=2\tag{2}$$
Applying inequality $(2)$ to equation $(1)$, we have $$[ABCD]\ge\dfrac{1}{2}\Big{(}(2+2+2+2)\Big{)}=4$$
Therefore, the minimum area of the quadrilateral is $\boxed{4}$.
Note: It can be proved that the smallest area of any quadrilaterals intersecting all $4$ hyperbola branches of $x^2y^2=1$ is $4$. Combining the answer to this question, we can prove that the smallest area of any convex polygons intersecting all $4$ hyperbola branches of $x^2y^2=1$ is $4$, as any convex polygons with more than $4$ sides can be reduced to a smaller quadrilateral by removing a vertex.