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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 $2d_k<p$ 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?

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Finally, I found out that this problem was proposed on the Saint-Petersburg mathematical olympiad in 1988.

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.

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