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Proof of the Rearrangement Inequality 📂Geometry

Proof of the Rearrangement Inequality

Theorem 1

Suppose there is a plane regular simple closed curve $\alpha$ with length $L$.

If the area enclosed by $\alpha$ is denoted as $A$, then $$ L^{2} \ge 4 \pi A $$ In particular, the condition for $L^{2} = 4 \pi A$ is that $\alpha$ is a circle.

Description

In fact, the fact itself mentioned in this theorem is known to many people, whether intuitively or otherwise, because we encounter circles in numerous natural phenomena, even if we do not know the physical reason why water drops do not form sharp edges but instead form round shapes.

The name of the inequality, “isoperimetric,” means having a constant periphery, and the isoperimetric inequality itself answers the question, “When does the internal area become the largest when the periphery is constant as $L$?”

Proof

Part 0. Buildup

Consider two lines $l_{1} \parallel l_{2}$ that are parallel to the tangent of $\alpha$. Let’s consider the center $O$ of the circle $\beta$, which has a radius $r>0$ and touches both of these lines at the same time, and denote the direction parallel to $l_{1}, l_{2}$ as the $y$-axis and the perpendicular direction as the $x$-axis.

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Now to mark four points $A,B,C,D$, starting with $A,C$. $A = \alpha (0)$ is the point where $\alpha$ touches $l_{1}$, and $C = \alpha \left( s_{2} \right)$ is the point where $\alpha$ touches $l_{2}$. We intend to represent both curves with coordinates given a common parameter $s$. Here, we assume without loss of generality that $\alpha$ is a unit speed curve, which means $\left\| \alpha^{\prime}(s) \right\| = 1$. $$ \alpha (s) = \left( x(s) , y(s) \right) \\ \beta (s) = \left( z(s) , w(s) \right) $$ Then, the coordinates of the circle $\beta$ can be expressed as follows, based on the point $s_{2}$. $$ \begin{align*} z(s) =& x(s) \\ w(s) =& \begin{cases} - \sqrt{ r^{2} - x^{2} } & , \text{if } 0 \le s \le s_{2} \\ \sqrt{ r^{2} - x^{2} } & , \text{if } s_{2} \le s \le L \end{cases} \end{align*} $$ Be careful not to confuse the $x$-axis with $x(s)$, and the $y$-axis with $y(s)$. When dealing with $x,y$ as vectors, always interpret it as $\left( x(s),y(s) \right)$, which represents the coordinates of $\alpha$.


  1. Millman. (1977). Elements of Differential Geometry: p64. ↩︎