Inequality for convex polygonsLet P be a convex n-gon on the plane. For k = 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 &lt; a 1 d 1 + a 2 d 2 + … + a n d n ≤ 4.Firstly, let us prove the first inequality. Indeed, if p is the perimeter of the polygon P, then it's clear that 2 d k &lt; p for all k ∈ { 1 , 2 , … , n }. Hence, we obtain a 1 d 1 + a 2 d 2 + … + a n d n &gt; 2 ( a 1 + a 2 + … + a n ) p = 2 p 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 ) := a 1 d 1 + a 2 d 2 + … + 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.So, is there any way to end this solution?

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Inequality for convex polygonsLet P be a convex n-gon on the plane. For   k  =            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  &amp;lt;            a      1              d      1        +            a      2              d      2        +  …  +            a      n              d      n        ≤  4.Firstly, let us prove the first inequality. Indeed, if p is the perimeter of the polygon P, then it&#039;s clear that   2      d    k    &amp;lt;  p for all   k  ∈  {  1  ,  2  ,  …  ,  n  }. Hence, we obtain            a      1              d      1        +            a      2              d      2        +  …  +            a      n              d      n        &amp;gt;            2      (              a        1            +              a        2            +      …      +              a        n            )        p    =            2      p        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  )  :=            a      1              d      1        +            a      2              d      2        +  …  +            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&#039;s sufficient to prove the inequality for Q, i. e. for centrally symmetric polygons (it&#039;s well-known that   P  +      P    ′   is a centrally symmetric polygon). However, it&#039;s quite unclear how to continue this approach.So, is there any way to end this solution?

Nkgopotsev1g · Accepted Answer

Step 1As was mentioned before it&#039;s sufficient to prove the inequality for centrally symmetric polygons. So, suppose that   n  =  2  m and   P  =      A    1        A    2    …      A          2      m       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    k        h    k    ≥  S for every   k  ∈  {  1  ,  2  ,  …  ,  2  m  }.Step 2Hence,       ∑          k      =      1              2      m                  a      k              d      k        ≤      ∑          k      =      1              2      m                          a        k                    h        k              S    .Now note that       a    k        h    k   is equal to the area of parallelogram       B    i        B          i      +      1            B          i      +      m            B          i      +      m      +      1      , but   S  (      B    i        B          i      +      1            B          i      +      m            B          i      +      m      +      1        )  =  4  S  (  O      B    i        B          i      +      1        ) (the corresponding four triangles has the same area). Thus,       ∑          k      =      1              2      m                  a      k              d      k        ≤      ∑          k      =      1              2      m                          a        k                    h        k              S    =      ∑          k      =      1              2      m                  4      S      (      O              B        i                    B                  i          +          1                    )        S    =            4      S        S    =  4  ,

Let P be a convex n-gon on the plane. For k=bar{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<a_1/d_1+a_2/d_2+...+a_n/d_n leq 4.

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