INTERNATIONAL JOURNAL OF GEOMETRY Vol. 7 (018), No. 1, 81-86 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 + 4 + 1 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: 06.08.017. In revised form: 15.0.018. Acceted: 5.0.018.
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 8 5 1 + t + + 1 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:
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)
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 4 1 + : 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, 4 1 + abcd This comletes the roof of Lemma 1..
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: 4 1 + 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.
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), 96-106. [4] Bottema, O., Geometric Inequalities, Wolters-Noordho Publishing, The Netherlands, 1969, 19-1. [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), 17-1. [7] Josefsson, M., A few inequalities in quadrilateral, International Journal of Geometry, 4 (1) (015), 11-15. [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), 10-11. [10] Stoica, E., Minculete, N. and Barbu, C., New asects of Ionescu Weitzenbock s inequality, Balkan Journal of Geometry and Its Alications, 1 () 016, 95-101. DROBETA TURNU SEVERIN TRAIAN NATIONAL COLLEGE, ROMANIA E-mail address: leonardgiugiuc@yahoo.com KIEN XUONG, THAI BINH, VIET NAM E-mail address: daothanhoai@hotmail.com EMIRDAG ANADOLU LISESI EMIRDAG, AFYON, TURKEY E-mail address: kadiraltintas1977@gmail.com