# Inequality for convex polygons

I have the following problem:

Problem. Let $$P$$ be a convex $$n$$-gon on the plane. For $$k=\overline{1,n}$$ define $$a_k$$ as the length of $$k$$-th side of $$P$$ and $$d_k$$ as the length of projection of $$P$$ onto the line containing $$k$$-th side of the polygon $$P$$. Prove that $$2<\frac{a_1}{d_1}+\frac{a_2}{d_2}+\ldots+\frac{a_n}{d_n}\leq 4.$$

Firstly, let us prove the first inequality. Indeed, if $$p$$ is the perimeter of the polygon $$P$$, then it's clear that $$2d_k for all $$k\in\{1,2,\ldots,n\}$$. Hence, we obtain $$\frac{a_1}{d_1}+\frac{a_2}{d_2}+\ldots+\frac{a_n}{d_n}>\frac{2(a_1+a_2+\ldots+a_n)}{p}=\frac{2p}{p}=2,$$ as desired.

Now, for the second part note that equality holds if, for example, $$P$$ is a rectangle, so the second inequality is sharp. For polygon $$P$$ denote $$f(P):=\frac{a_1}{d_1}+\frac{a_2}{d_2}+\ldots+\frac{a_n}{d_n}.$$ Then, it can be shown that if $$P'$$ is polygon which is centrally symmetric to $$P$$, then the Minkowski sum $$Q=P+P'$$ satisfy the following equality $$f(Q)=f(P).$$ Thus, it's sufficient to prove the inequality for $$Q$$, i. e. for centrally symmetric polygons (it's well-known that $$P+P'$$ is a centrally symmetric polygon). However, it's quite unclear how to continue this approach (I don't even sure that problem became easier).

So, is there any way to end this solution?

As was mentioned before it's sufficient to prove the inequality for centrally symmetric polygons. So, suppose that $$n=2m$$ and $$P=A_1A_2\ldots A_{2m}$$ and let $$O$$ be center of the symmetry of $$P$$. Let $$h_k$$ be the length of projection of $$P$$ onto line which is perepndicular to $$k$$-th side of polygon $$P$$. Denote the area of polygon $$P$$ as $$S$$. Then, due to convexity of $$P$$ we have $$d_kh_k\geq S$$ for every $$k\in\{1,2,\ldots,2m\}$$. Hence, $$\sum_{k=1}^{2m}\frac{a_k}{d_k}\leq\sum_{k=1}^{2m}\frac{a_kh_k}{S}.$$ Now note that $$a_kh_k$$ is equal to the area of parallelogram $$B_iB_{i+1}B_{i+m}B_{i+m+1}$$, but $$S(B_iB_{i+1}B_{i+m}B_{i+m+1})=4S(OB_iB_{i+1})$$ (the corresponding four triangles has the same area). Thus, $$\sum_{k=1}^{2m}\frac{a_k}{d_k}\leq\sum_{k=1}^{2m}\frac{a_kh_k}{S}=\sum_{k=1}^{2m}\frac{4S(OB_iB_{i+1})}{S} =\frac{4S}{S}=4,$$ as desired.