Some Inequalities for Convex Sets

: The paper concerns inequalities between fundamental quantities as area, perimeter, diameter and width for convex plane fugures.


Introduction
In this paper we use methods from the geometry of convex figures and geometric inequalities. Both the above mathematical subjects are old but have been very convenient, fruitful and active in recent times. A classical textbook in the convexity is the excellent, "Theory of Convex Bodies" by T. Bonnesen and W. Fenchel [1]. Another very nice tool is the "Convex Figures" of I. Yaglom and V. Boltyanski [2]. The Yugoslavian and Romanian school, with Mitvinovic and Andrescu, produced a very interesting theorem in algebraic and geometric Inequalities. Here, the problem is to find solutions to three interesting inequalities for convex figures in the plane. This problem came to me from a former student of mine, Prof. E. Symeonidis. The problem was published in AXIOMS 2018 7(1) by S.Marcus and F. Nichita. The solution of the first inequality is quite simple but the two others are sophisticated. For the solution I used two lemmas that have particular interest and could be useful to solve other problems.

Problems
Let f be a convex figure in the plane (that is a compact convex set). We denote by G its centroid. D is the maximal chord and d the minimal chord through G. Moreover, L stands for the perimeter, D F the diameter, d F the minimal breadth, and A the area of F.
We have to prove:

Proof. Inequality (a).
The formula for the perimeter of a convex figure F see [3] is: where B(ϑ) is the breadth of F to the direction ϑ, d F is the min. breadth of F. So we have: This is because of the obvious d F ≥ d The equality holds for the circle and the convex figures with constant breadth.
For (b) and (c) we need two lemmas.

Lemma 1.
Let F be a convex figure and AB a diametrical chord. We denote by l 1 , l 2 the support lines at the points A, B respectively. We take the chord CD parallel to AB, so that CD = AB/2. The str.line CD intersects l 1 , l 2 at the points K, L. The chord AB disects F into two parts. We denote the one part by F 1 as in the Figure 1. Then we have : area(ABLK) ≥ areaF 1 .
Proof. Let AC intersect BD at the point M and MS parallel to AK, The support line at the point C intrsects AK, MS at the points p, q. The support line at the point D intersects BL, MS at the points p , q . In the triangle AMB the points C and D are the midle points of the sides, so we use equalities of triangles without the proofs. The triangles pKC, qSC are equal, the same for ApC, MqC, the same for LDp , SDq and Mq D, Bp D. We denote w 1 , w 2 , w 0 the area of the segments of the arcAC, arcBD, arcCD.
We have By w 1 we denote the area of the triangle ApC with sides Ap, pC, arcAC. Analogously w 2 . Hence we conclude Area(ABLK) > AreaF 1 . Proof. We denote by M the vector OM.

We have
and From the above (1), (2) follows From (3) and (4) we take and finally The continuity of the convexity asserts us that we can choose the diametrical chord AB so that: where d F and D F stands for the min.breadth and diameter of F respectively. As you see in Figure 3, l 1 and l 2 are the parallel supporting lines at the points A and B.
We take the points D, C, E, F on θF so that CD = EF = AB 2 , and CD EF AB.
We easily see, according our first lemma that where h is the distance between CD, EF. We now see that the convex F and the orthogonal PQLK have common part F ∩ PQLK; therefore, the position of the centroid G of F depends on the centroid g of the segments arcCD and segment arcEF. The point g lies inside the orthogonal PQLK. So we conclude that d ≥ h and from (6) follows that

Inequality (c).
The equality only for F circle. We suppose that F is not a circle. We translate the str. lines KL, PQ closer towards to AB until to K L , P Q such a way the parallelogram K L Q P has Area(K L Q P ) = Area(F) we have: Let now L 1 be the part of the perimeter L over of AB and analogously L 2 . We see that C D /AB > 1/2 and E F /AB > 1/2. So, from the lemma 2, we easily see that arcC D > L 1 /2 and arcE F > L 2 /2. That is arcC D + arcE F > L/2 > arcE C + arcD F . Moreover, AreaF < lengtharcE C .AB , AreaF < lengtharcD F. AB; therefore, 2AreaF < (lengtharcE C + lengtharcD F )AB but lengtharcE C + lengtharcD F < lengtharcC D + lengtharcE F ; hence, 4AreaF < (lengtharcE C + lengtharcD F )AB + (lengtharcC D + lengtharcE F)AB < AB.L < D.L.
Funding: This research received no external funding.