AN INEQUALITY RELATED TO THE LENGTHS AND AREA OF A CONVEX QUADRILATERAL

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1 INTERNATIONAL JOURNAL OF GEOMETRY Vol. 7 (018), No. 1, AN INEQUALITY RELATED TO THE LENGTHS AND AREA OF A CONVEX QUADRILATERAL LEONARD MIHAI GIUGIUC, DAO THANH OAI and KADIR ALTINTAS Abstract. In this aer we give an inequality related to the lengths and the area of a convex quadrilateral and its roof. 1. Introduction There are many famous inequalities related to the lengths and area of a triangle, for examles with a triangle we have Pedoe s inequality [9], Weitzenbock s inequality [10], Ono s inequality [8], Blundon s inequality []. In a quadrilateral we have many imortant inequalities, such as Yun s inequality [6], Josefsson s inequality [6],[7]; some other inequalities of a quadrilateral you can see in [4], [1], []. In this aer we give a nice inequality related to the lengths and area of a convex quadrilateral in theorem 1.1 as follows: Theorem 1.1. Let a; b; c; d be the lengths of the sides of a convex quadriateral ABCD with the area S, the following inequality hold: (1) 1 + (ab + ac + ad + bc + bd + cd) 1 ( + 1) (a + b + c + d ) S Equality hold if only if ABCD is a square. To rove Theorem 1.1, rst we give a stronger inequality as follows: Theorem 1.. If a; b; c and d be the lengths of the sides of a convex quadrilateral ABCD, the following inequality hold: () (a + b + c + d) 5 a + b + c + d abcd Equality hold if only if a = b = c = d. Keywords and hrases: geometry inequality, cyclic olygon, (010)Mathematics Subject Classi cation: 51M04, 51M99 Received: In revised form: Acceted:

2 8 Leonard Mihai Giugiuc, Dao Thanh Oai and Kadir Altintas Proof. Figure 1 In order to rove (), we need to establish few things rst. Since ABCD is the convex quadrilateral, we have 0 < a; b; c; d < a+b+c+d : Because both members of (1) are homogenous of second degree, we may assume a + b + c + d = 4. Thus, 0 < a; b; c; d <. By Maclaurin s inequality we have: i.e. ab + ac + ad + bc + bd + cd 6 So that there exist t [0; 1) such that: a + b + c + d 4 ab + ac + ad + bc + bd + cd 6: () ab + ac + ad + bc + bd + cd = 6(1 t ): From () a + b + c + d = (a + b + c + d) (ab + ac + ad + bc + bd + cd) = 16 1(1 t ) = 4(1 + t ): Thus (1) is equivalent to t abcd, ( + 1) (5 )t + 1 abcd, (4) 1 4 t abcd: De ne the olynomial P : (0; 1)! R as P (x) = (x a)(x b)(x c)(x d); 8x > 0: By Viete s theorem combined with (), we have: P (x) = x 4 4x + 6(1 t )x 4sx + ; 8x > 0, where 4s = abc + abd + acd + bcd and = abcd:

3 An inequality related to the lengths and area of a convex quadrilateral 8 Lemma 1.1. Let a; b; c and d be the lengths of the sides of a convex quadrilateral ABCD with erimeter equal to 4 then (5) a + b + c + d < 8: Proof. Denote u = a 1, v = b 1, w = c 1 and z = d 1. Then 1 < u; v; w; z < 1 and u + v + w + z = 0. The inequality (5) is equivalent with u + v + w + z < 4, which is true since 0 u ; v ; w ; z < 1. This comletes the roof of Lemma 1.1. Note that Lemma 1.1 combined to (4) gives us that t [0; 1 ): Lemma 1.. If 0 t < 1 then (1 t) (1 + t): Proof. Consider the function f : (0; 1)! R; f(x) = P (x) x. Since f admits four ositive roots, the from Rolle s theorem, f 0 admits at least ositive roots. But f 0 (x) = x4 8x + 6(1 t )x ) x ; 8x > 0: From the obove consideration, the olynomial g(x) = x 4 8x + 6(1 t )x admits at least three roots in (0; 1). We have: thus the critical oints of g are 1 g 0 (x) = 1x x x + (1 t ) ; t 1 + t and f. We form the function g the Rolle s sequence 0 + < 1 t 1 + t < 1. Since f(0 + ) = 1 < 0; f(1) = +1 > 0 and f has three ositive roots, the from the consequence of Rolle s theorem, we obtain This comletes the roof of Lemma 1.. f(1 t) 0 ) (1 t) (1 + t): Lemma 1.. Let a; b; c and d be real numbers situated in the interval [0; ] such that a+b+c+d = 4 and ab + ac + ad + bc + bd + cd = 6(1 t ). If 1 t 1 then: 4 (6) 1 + (9t 1) Moreover = 4 1+ i (a; b; c; d) = and ermutations. : ; ; ; + Proof. Let k be a real number with 0 < k <. We consider the real numbers a; b; c and d situated in the interval [0; ] such that a + b + c + d = 4 and a + b + c + d = (k 4k + 4). Then, abcd 4k (1 k): First let s remark that if (a; b; c; d) = (k; k; k; ) and their ermutations, then a + b + c + d = 4; a + b + c + d = (k 4k + 4)

4 84 Leonard Mihai Giugiuc, Dao Thanh Oai and Kadir Altintas and abcd = 4k (1 k). Let ; be ositive real numbers with >. We consider the non negative real numbers u; v; w and t such u + v + w + t = + and u + v + w + t = +. Then uvw + uvt + uwt + vwt : The result is known, hence we don t need to rove it here. Back to the roblem. Putting = k, = k, u = a, v = b, w = c and t = d we get: > > 0; u; v; w; t 0; u + v + w + t = + and u + v + w + t = + : Hence according to the lemma, uvw + uvt + uwt + vwt ) abc + abd + acd + bcd k( k k + 4): We assume by absurd that there exist a; b; c; d [0; ] with a+b+c+d = 4 and a +b +c +d = (k 4k + 4) such that abcd > 4k (1 k): De ne the olynomials f; g : [0; ]! R, as f(x) = (x a)(x b)(x c)(x d) and g(x) = (x k) (x (1 k)) (x ), 8x [0; ]. Then f(x) = x 4 4x + ( k + 4k + 4)x mx + ; 8x [0; ] where m = abc + abd + acd + bcd, and = abcd and g(x) = x 4 x + ( k + 4k + 4)x k( k k + 4)x + 4k (1 k); 8x [0; ]: Thus, f(x) g(x) = k( k k + 4) m x + 4k (1 k) > 0, 8x [0; ]. Hence, if fa; b; c; dg, then f() g() > 0 ) g() < 0. But g(x) < 0 if and only if x ((1 k); ) ) fa; b; c; dg ((1 k); ). Consequently, via Rolle s theorem, the roots of f 0 are contained in the interval ((1 those roots. Then which is false, because k < (1 abcd 4k (1 f 0 (x) = 4(x y 1 )(x y )(x y ) 0; k); ). Let y 1 ; y and y be k) < y 1 ; y ; y. So that our assumtion was false. In conclusion, k). Back to the lemma s 1. roof. If 1 < t < 1, let s remark rst that the exists k (0; ) such that a + b + c + d = (k 4k + 4). Indeed, solving the quadratic (k 4k +4) = 4(1+t ), we choose the root k = and clearly 0 < k <. Hence abcd : If t = 1, then, by the roof of the lemma 1., abcd (1 t) (1 + t). If t = 1, the only choices we have are (a; b; c; d) = (; ; 0; 0) and ermutations and clearly, abcd This comletes the roof of Lemma 1..

5 An inequality related to the lengths and area of a convex quadrilateral 85 We are now ready to rove the inequality (): Case 1: 0 t 1 : By the lemma we have: (1 t) (1 + t), hence su ce it to show: 1 ( 4)t (1 t) (1 + t) (11 6 )t 0 t (1 t) which is true since 0 t < 1 and 11 6 > 1 : We deduce from the above roof that in this case equality holds at t = 0, a = b = c = d: Case : 1 t 1 : The fact that 0 < a; b; c; d < combined with the Lemma 1. give us that: Su ce it to show that: 1 ( v u t4 4)t 1 + Which is true after strightward calculations. This comletes the roof of Theorem 1... PROOF OF THE THEOREM 1.1 By famous Brahmaguta s formula and Bretschneider s formula (see [5]), we know that among all convex quadrilaterals ABCD with the lengths of the side a; b; c and d then the cyclic ones have the lagest area. So we only need to rove the Theorem 1.1 with ABCD is the convex cyclic quadrilateral. By the Brahmaguta s formula we have that the area of convex cyclic quadrilateral ABCD is S = (s a)(s b)(s c)(s d) where s = a+b+c+d. So the inequality (1), 4 (ab+ac+ad+bc+bd+cd) ( 1)(a +b +c +d )+4( +1) (s a)(s b)(s c)(s d), (7) (a + b + c + d) (5 )(a + b + c + d ) + 4 ( + 1) (s a)(s b)(s c)(s d) As above, without loss of generality, we may assume that a + b + c + d = 4. Denote x = a, y = b; z = c and w = d. Then 0 < x; y; z; w <. Moreover, a = x, b = y, c = z, d = w and x + y + z + w = 8 (a + b + c + d) = 4 so that x; y; z and w are the lengths of the sides of a covex quadrilateral XY ZW. Noted that a + b + c + d = ( x) + ( y) + ( z) + ( w) = 16 4(x + y + z + w) + x + y + z + w = x + y + z + w. Thus, (7) be come: (8) (x + y + z + w) (5 )(x + y + z + w ) + 4 ( + 1) xyzw Equality in (8) holds if only if a = b = c = d and ABCD is a cyclic quadrilateral. The inequality (8) is true according to the (). This comlete the roof of Theorem 1.1.

6 86 Leonard Mihai Giugiuc, Dao Thanh Oai and Kadir Altintas The authors would like to thank the referees for their valuable comments which heled to imrove the manuscrit. References [1] Alsina, C. and Nelsen, R., When Less is More: Visualizing Basic Inequalities, Mathematical Association of America, 009. [] Andreescu, T., Andrica, D., 60 Problems for Mathematical Contests, GIL Publishing House, 00. [] Andrica, D., Barbu, C. and Minculete, N., A Geometric way to generate Blundon Tye inequalities, Acta Universitatis Aulensis, 1 (01), [4] Bottema, O., Geometric Inequalities, Wolters-Noordho Publishing, The Netherlands, 1969, [5] Brahmaguta s formula, htts://en.wikiedia.org/wiki/brahmagutans_formula. [6] Josefsson, M., Five Proofs of an Area Characterization of Rectangles, Forum Geometricorum, 1 (01), [7] Josefsson, M., A few inequalities in quadrilateral, International Journal of Geometry, 4 (1) (015), [8] Ono, T., Problem 4417, Intermed. Math., 1 (1914), 146. [9] Pedoe, D., An Inequality Connecting Any Two Triangles, The Mathematical Gazette, 5 (67) (1941), [10] Stoica, E., Minculete, N. and Barbu, C., New asects of Ionescu Weitzenbock s inequality, Balkan Journal of Geometry and Its Alications, 1 () 016, DROBETA TURNU SEVERIN TRAIAN NATIONAL COLLEGE, ROMANIA address: leonardgiugiuc@yahoo.com KIEN XUONG, THAI BINH, VIET NAM address: daothanhoai@hotmail.com EMIRDAG ANADOLU LISESI EMIRDAG, AFYON, TURKEY address: kadiraltintas1977@gmail.com