Example 5: It is desired to cut out a triangle of arbitrary shape with an area of 1 from an infinitely long strip of paper. What is the minimum width of the strip?

Analysis: Since it is necessary to estimate the width of the strip, let's start with special triangles first.

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Prove: For an equilateral triangle with an area of 1, the side length is $\frac{2}{\sqrt{3}}$, and the height is $\sqrt[4]{3}$. Therefore, it is conjectured that the width of the paper strip should be at least $\sqrt[4]{3}$. Assume that an equilateral $\triangle ABC$ with an area of 1 can be cut from a paper strip narrower than $\sqrt[4]{3}$, as shown in Figure 1. Now project the three vertices of $\triangle ABC$ to the edges of the paper strip at points $A_{1} 、 B_{1} 、 C_{1}$ (with $B_{1}$ between $A_{1} 、 C_{1}$). Connect $B B_{1}$ and let it intersect $A C$ at point $M$. Then, the height of $\triangle ABC$ does not exceed $B M$. Since $B M$ does not exceed the width of the paper strip, the height of $\triangle ABC$ is less than $\sqrt[4]{3}$. Therefore, the area of $\triangle ABC$ is less than 1, which is a contradiction.

So, is a paper width of $\sqrt[5]{3}$ sufficient? Assume it is not, which means the heights of the triangle with an area of 1 all exceed $\sqrt[4]{3}$, and thus, the side lengths $a 、 b 、 c$ are all less than $\frac{2}{\sqrt[4]{3}}$. Let the smallest angle of $\triangle ABC$ be $\angle C$, and $\angle C \leqslant 60^{\circ}$. Therefore, $S_{\triangle ABC}=\frac{1}{2} a b \sin C < 1$, which is a contradiction.
In conclusion, the width of the paper strip should be at least $\sqrt[4]{3}$.