J ournal of Mathematical I nequalities Volume 7, Number (2013), 25–32 doi:10.7153/jmi-07-03 ´ SOME GEOMETRIC INEQUALITIES OF RADON –– MITRINOVIC ˘ ¸ U -G IURGIU AND N ECULAI S TANCIU D M B ATINET (Communicated by J Peˇcari´c) Abstract Some Mitrinovi´c type inequalities for general convex polygons are presented The main tool in the proofs is the Radon inequality Introduction Inequality of J Radon has the statement: n n k=1 k=1 If n ∈ N ∗ , m ∈ R+ , xk ∈ R+ , yk ∈ R∗+ , ∀k = 1, n , Xn = ∑ xk , Yn = ∑ yk , then: xm+1 k m k=1 yk n ∑ Xnm+1 Ynm (R) Equality holds if and only if there exists t ∈ R∗+ such that xk = tyk , ∀k = 1, n Inequality of Dragoslav S Mitrinovi´c has the statement: In any triangle with perimeter 2p , circumscribed of circle C(I; r) occurs the inequality √ p 3r (M) Equality holds if and only if the triangle is equilateral Results The purpose of this article is to establish some geometric inequalities (other then [1]) on (M)-type, in convex polygons, used the (R)-inequality For any convex polygon A1 A2 An , n we denoted by S the area of the polygon, by 2p the perimeter of the polygon, by ak the side of the length [Ak Ak+1 ] k = 1, n , An+1 ≡ A1 , and for any point M from inside the polygon we denoted Tk = prAK AK+1 M , dk = MTk , Sk = area [Ak MAk+1 ], ∀k = 1, n, and uk = μ (∠Ak MAk+1 ) , vk = μ (∠Ak MTk ) , wk = μ (∠Tk MAk+1 ) the measures of angles ∠Ak MAk+1 , ∠Ak MTk , respectively ∠Tk MAk+1 in radians Mathematics subject classification (2010): 26D15, 26D05, 51M16 Keywords and phrases: Mitrinovi´c type inequality, Radon’s inequality, convex polygon, convex polyhedron c Ð , Zagreb Paper JMI-07-03 25 26 ˘ D M B ATINET ¸ U -G IURGIU AND N S TANCIU L EMMA If A, B, A = B are points in the same plane and M ∈ / AB, T = prAB M , d = MT , then: AB u AB = tg , d MT where u = μ (∠AMB) is the measure of the angle ∠AMB in radians Proof We denoted v = μ (∠AMT ) and w = μ (∠T MB) We have the following possibilities: i) T ∈ (AB) In this case tg v = AT MT and tg w = TB MT , so AB AT + T B AB = = = tg v + tg w d MT MT Since the function f : 0, π2 → R+ , f (x) = tg x , is convex on 0, π2 Jensen’s inequality that AB = tg v + tg w d tg AB d , and tg u = 1− tg 22 u ii) T ≡ A In this case tg w = tg u = tg w, so AB d = tg u = tg u it follows by v+w u = tg 2 tg v = u AT d tg ⇔ TT d u tg = = Also we have: , which is true iii) T ≡ B.This case is analogous with ii).Hence the conclusion follows TA MT iv) A ∈ (T B) We have tg v = Hence, = TA d and tg w = tg (u + v) = AB T B − TA = = tg (u + v) − tg v tg u d d tg u + tg v ⇔ − tg u − tg v ⇔ tg u tg v( tg u + tg v) − tg u tg v which is true Therefore, AB d tg u TB MT = TB d 0, tg u2 v) B ∈ (AT ) This case is analogous with iv) Hence the conclusion follows And we are done T HEOREM If A1 A2 An (n this polygon, then: 3) is a convex polygon and M is a point inside n ak ∑ dk k=1 2n tg π n Proof By lemma we have: ak dk tg uk , ∀k = 1, n (1) S OME GEOMETRIC INEQUALITIES OF R ADON – M ITRINOVI C´ and then n n ak ∑ dk ∑ tg k=1 Since the function f : 0, inequality follows that: n uk ∑ tg π k=1 uk → R+ , f (x) = tg 2x , is convex on 0, π2 n ∑ uk 2n k=1 n tg k=1 27 = n tg by Jensen’s 2π π = n tg 2n n O BSERVATION 1.1 If the polygon A1 A2 An is circumscribed in the circle C(I; r) and M ≡ I , then: dk = r , ∀k = 1, n, and (1) becomes: n 2p ak = ∑ r k=1 r 2n tg π ⇔p n nr tg π n (1 ) O BSERVATION 1.2 The inequality (1) is a generalization of the (M)-inequality, and if the polygon is the triangle ABC (1) becomes: b c a + + da db dc tg √ π =6 3 (1 ) and if M ≡ I , then we obtain the (M)-inequality T HEOREM If A1 A2 An (n polygon and m ∈ R+ , then: 3) is a convex polygon, M is a point inside this n ak ∑ dm k=1 k 2· pm+1 Sm n am+1 (2) Proof We have: n ak am+1 n am+1 k m , S k k=1 n ∑ d m = ∑ (akkdk )m = ∑ 2mk Sm = 2m ∑ k=1 k k=1 k=1 k and used the (R)-inequality, we obtain n a ∑ d mk k=1 k 2m n m+1 k=1 n m ∑ ak ∑ Sk = 2m+1 pm+1 pm+1 · = · 2m Sm Sm k=1 O BSERVATION 2.1 If the polygon A1 A2 An (n circle C(I; r), then S = pr , and (2) becomes: n ak ∑ dm k=1 k 2· pm+1 2p = m pm r m r 3) is circumscribed in the (2 ) 28 ˘ D M B ATINET ¸ U -G IURGIU AND N S TANCIU If M ≡ I then by (1 ), we deduce that: n 2n π tg rm−1 n ak ∑ dm k=1 k T HEOREM If A1 A2 An (n polygon, and m ∈ R+ , then: n 3) is a convex polygon, M is a point inside this 2m nm+1 m+1 π tg p n ak ∑ d m+1 k=1 k n Proof We have: ∑ k=1 n ak and we obtain ∑ m+1 k=1 dk ak dkm+1 n a ∑ dk k=1 k n n = ∑ k=1 m+1 ak dk am k = n 2p ∑ k=1 k=1 (3) , where we apply the (R)-inequality m+1 ∑ ak conclusion (2 ) m+1 ak dk , then by (1) we deduce the O BSERVATION 3.1 The relation (3) is also written as: n ak m+1 k=1 dk p∑ 2m nm+1 tg m+1 π n (3 ) If the polygon A1 A2 An (n 3) is circumscribed in the circle C(I; r), then (3 ) becomes: π π m+1 (3 ) 2p2 · m+1 2m nm+1 tg m+1 ⇔ p2 2m−1 nr tg r n n T HEOREM If A1 A2 An (n 3) is a convex polygon, M is a point inside this polygon and m ∈ R+ , then for any x, y ∈ R∗+ such that 2px > y max ak , occurs the k n inequality: n nm+1 ak ∑ d m (2px − ya )m+1 (4) 2m Sm (nx − y)m+1 k k=1 k Proof We have: n am+1 n ak n ∑ d m (2px − ya )m+1 = ∑ am d m (2pxk − a y)m+1 = 2m ∑ Sm k k=1 k k=1 k k=1 k k k ak 2px − yak m+1 , where we apply the (R)-inequality and we obtain that n n ∑ ak m+1 m k=1 dk (2px − yak ) · 2m ∑ k=1 m+1 ak 2px−yak n ∑ Sk k=1 m = m m S n ak ∑ 2px − yak k=1 m+1 29 S OME GEOMETRIC INEQUALITIES OF R ADON – M ITRINOVI C´ Also we have, Un = n n ak yak ∑ 2px − yak ⇔ yUn = ∑ 2px − yak k=1 k=1 ⇔ yUn + n = n n yak + = 2px ∑ 2px − yak k=1 2px − yak ∑ k=1 By Bergstrăoms inequality we deduce that: n n2 2px − yak n ∑ (2px − yak ) k=1 = n2 2p (nx − y) k=1 Therefore, n 2px − yak k=1 yUn + n = 2px ∑ ny n2 x −n = ⇔ Un nx − y nx − y ⇔ yUn Hence, 2px · n n nx − y nm+1 ak ∑ d m (2px − ya )m+1 2m Sm (nx − y)m+1 k k=1 k n2 x n2 = 2p(nx − y) nx − y O BSERVATION 4.1 If the polygon from theorem is circumscribed in the circle C(I; r), then for any m ∈ R+ and for any point M which is inside the polygon holds the inequality: n nm+1 ak ∑ d m (2px − ya )m+1 2m Sm (nx − y)m+1 k k=1 k = nm+1 2m pm rm (nx − y)m+1 (4 ) If M ≡ I , then n nm+1 ak ∑ rm (2px − ya )m+1 k k=1 2m pm rm (nx − y)m+1 n ak nm+1 k=1 (2px − yak)m+1 2m (nx − y)m+1 ⇔pm ∑ (4”) O BSERVATION 4.2 If the polygon from theorem is the triangle ABC , then (4) becomes: a m+1 dam (2px − ay) + b dbm (2px − by)m+1 + 3m+1 c m+1 dcm (2px − cy) 2m Sm (3x − y)m+1 (4 ) 30 ˘ D M B ATINET ¸ U -G IURGIU AND N S TANCIU Also (4 ) becomes: pm a (2px − ya)m+1 + b (2px − yb)m+1 + c 3m+1 (2px − yc)m+1 2m (3x − y)m+1 If we take m = , x = y = , then by the last relation we obtain: a b c + + b+c c+a a+b , and this is Nesbitt’s inequality T HEOREM If a convex polyhedron has n faces ( n ) which are convex polygons with their areas Sk k = 1, n , M is a point inside the polyhedron, dk is the distance from M to the face with area Sk k = 1, n , V is the volume of the polyhedron, S is total area of the polyhedron and m ∈ R+ , then: n Sm+1 3mV m Sk ∑ dm k=1 k (5) Proof We have: n n Sm+1 Skm+1 Sk = = ∑ d m ∑ (dk Sk )m ∑ 3mk V m , k k=1 k k=1 k=1 n where Vk is the volume of the pyramid which has the vertex M and which has the base the polygon of the face with the area Sk k = 1, n By (R)-inequality, we deduce that: m+1 n n ∑ Sk Sk k=1 ∑ dm k=1 k m n 3m ∑ Vk = Sm+1 3mV m k=1 O BSERVATION 5.1 If the convex polyhedron from theorem is circumscribed in the sphere S(I; r), then (5) becomes: n Sk ∑ dm k=1 k Sm+1 S = m (Sr)m r (5 ) If M ≡ I , then dk = r , ∀k = 1, n , and the inequality (5 ) becomes a equality T HEOREM If we have a convex polyhedron like in theorem and m ∈ R+ , x, y ∈ R∗+ , such that xS y max Sk , then: k n n Sk ∑ d m (xS − yS )m+1 k=1 k k nm+1 3mV m (nx − y)m+1 (6) 31 S OME GEOMETRIC INEQUALITIES OF R ADON – M ITRINOVI C´ Proof We have: n Sm+1 n Sk n ∑ d m (xS − yS )m+1 = ∑ (d S )m (xSk − yS )m+1 = ∑ 3mV m k k=1 k k k k=1 k k k=1 Sk xS − ySk m+1 , where we apply the (R)-inequality and we obtain that: n n ∑ Sk ∑ k=1 m+1 m k=1 dk (xS − ySk ) m+1 Sk xS−ySk m n 3m 3mV m = ∑ Vk n Sk ∑ xS − ySk k=1 m+1 k=1 Also we have, Wn = n n Sk ySk ∑ xS − ySk ⇔ yWn = ∑ xS − ySk k=1 ⇔yWn + n = k=1 n n ySk + = xS ∑ xS − ySk xS − ySk k=1 k=1 By Bergstrăoms inequality we deduce that: n n2 ∑ xS − ySk n ∑ (xS − ySk ) k=1 = n2 (nx − y) S k=1 So, yWn + n Therefore, x· n2 ⇔ yWn nx − y n ny xn2 −n = ⇔ Wn nx − y nx − y Sk ∑ d m (xS − yS )m+1 k k=1 k n nx − y nm+1 3mV m (nx − y)m+1 O BSERVATION 6.1 If convex polyhedron from theorem is circumscribed in the sphere S(I; r), then the relation (6) becomes: n nm+1 Sk ∑ m+1 m k=1 dk (xS − ySk ) Sm rm (nx − y)m+1 (6 ) If in addition M ≡ I , then: n Sk ∑ (xS − yS )m+1 k=1 k nm+1 Sm (nx − y)m+1 (6 ) 32 ˘ D M B ATINET ¸ U -G IURGIU AND N S TANCIU REFERENCES [1] M D INC A˘ AND M B ENCZE, Trip in world of geometrical inequalities (2), Octogon Mathematical Magazine 11, (April, 2003), 45–76 (Received June 17, 2011) D M B˘atinet¸u- Giurgiu Department of Mathematics “Matei Basarab” National College Bucharest, Romania Neculai Stanciu Department of Mathematics “George Emil Palade” Secondary School Buz˘au, Romania e-mail: stanciuneculai@yahoo.com Journal of Mathematical Inequalities www.ele-math.com jmi@ele-math.com ... ENCZE, Trip in world of geometrical inequalities (2), Octogon Mathematical Magazine 11, (April, 2 003) , 45–76 (Received June 17, 2011) D M B˘atinet¸u- Giurgiu Department of Mathematics “Matei Basarab”... Romania e-mail: stanciuneculai@yahoo.com Journal of Mathematical Inequalities www.ele-math.com jmi@ ele-math.com