THE TEACHING OF MATHEMATICS 2013, Vol XVI, 1, pp 1–5 GENERALIZATIONS OF SOME REMARKABLE INEQUALITIES D.M Bˇ atinet¸u-Giurgiu, Neculai Stanciu Abstract In this paper, we present a generalization of Mitrinovi´ c’s inequality for polygons, and triangles, a generalization of J Radon’s inequality and a generalization of Nesbitt’s inequality The main tool in the proofs is the inequality of Jensen MathEduc Subject Classification: H34 MSC Subject Classification: 97H30 Key words and phrases: Jensen inequality; Mitrinovi c inequality; J Radon inequality; Bergstră om inequality; Nesbitt inequality Generalizations of Mitrinovi´ c’s inequality for convex polygons and triangles If A1 A2 An , n is a convex polygon, and M ∈ int(A1 A2 An ), with prAk Ak+1 M = Tk ∈ [Ak Ak+1 ] for k ∈ {1, 2, , n}, An+1 = A1 , then n k=1 Ak Ak+1 M Tk 2n tan π n Proof We first prove the following Lemma Let A, B, A = B be points in the plane and M ∈ / AB, T = prAB M Then AB = tan u + tan v, MT where u = µ(∠AM T ), v = µ(∠T M B) are the measures of the angles ∠AM T and ∠T M B Proof of the Lemma We have the following cases: AT BT AB i) T ∈ (AB) Then tan u = and tan v = , hence tan u + tan v = MT MT MT AT AA BT ii) T ≡ A Then tan u = = = and tan v = , hence tan u + MT MT MT AB tan v = Similarly if T ≡ B MT Ak Ak+1 From Lemma, we have = tan uk + tan vk , for k = 1, n, where uk = M Tk µ(∠Ak M Tk ), vk = µ(∠Tk M Ak+1 ) It follows that n k=1 Ak Ak+1 = M Tk n (tan uk + tan vk ) k=1 D.M Bˇ atinet¸u-Giurgiu, N Stanciu Since the function f : [0, π/2) → [0, +∞), f (x) = tan x is convex on [0, π/2), we can apply Jensen’s inequality and obtain that n k=1 n k=1 (uk Since, Ak Ak+1 M Tk 2n tan 2n n (uk + vk ) k=1 + vk ) = 2π, we deduce that n Ak Ak+1 M Tk k=1 2n tan 2π π = 2n tan , 2n n and we are done Remark 1.1 M ≡ I, we have n Ak Ak+1 = r k=1 If A1 A2 An is circumscribed about the circle C(I; r) and M Tk = r, k = 1, n, and the obtained inequality becomes 2s π 2n tan , wherefrom, r n π (1) s nr tan n The inequality (1) is a generalization of Mitrinovi´c’s inequality for triangles √ (M) s 3r Remark 1.2 If A1 A2 A3 is a triangle, then the obtained inequality becomes √ A1 A2 A2 A3 A3 A1 π + + tan = M T1 M T2 M T3 For M ≡ I, we obtain (M) For more results see [1] A generalization of J Radon’s inequality In what follows, we denote R+ = [0, +∞) and R∗+ = (0, +∞) Let a, b, c, d, xk , yk ∈ R∗+ , k = 1, n and Xn = m, p, q, s ∈ R+ , r ∈ [1, +∞) are such that cYns > d max1 n (2) k=1 r(m+1) + bxqk )m+1 xk (cYns − dyks )m ykm (aXnp k xk , Yn = k=1 If s n yk , k = 1, n, then (anq Xnp+r + bXnq+r )m+1 (cns − n n k=1 m(s+1) d)m Yn · n(m+1)(q+r−1)−ms Proof Denote uk = (aXnp +bxqk )xrk , vk = (cYns −dyks )yk , k = 1, n, Vn = and the left-hand side of (2) becomes n k=1 um+1 k = vkm n vk k=1 uk vk n m+1 = Vn k=1 vk Vn uk vk n k=1 vk m+1 Since the function f : R∗+ → R∗+ , f (x) = xm+1 is convex, we use Jensen’s inequality and we obtain that n k=1 vk f Vn uk vk n f k=1 vk u k · Vn v k n =f k=1 uk Vn = Vnm+1 n m+1 uk k=1 Generalizations of some remarkable inequalities Therefore, n k=1 um+1 k vkm n Vn m+1 uk Vnm+1 k=1 n = m Vn m+1 uk , k=1 i.e., n n r(m+1) + bxqk )m+1 xk (cYns − dyks )m ykm (aXnp k=1 n aXnp xrk + b k=1 n cYns yk k=1 = n k=1 n k=1 n −d k=1 aXnp = m yks+1 m (cYns − dyks )yk k=1 m+1 xq+r k m+1 (aXnp + bxqk )xrk n k=1 xrk + b cYns+1 −d n k=1 n k=1 m+1 xq+r k yks+1 m Since the functions g, h, k : R∗+ → R∗+ , g(x) = xr , h(x) = xq+r , k(y) = y s+1 are convex, also by Jensen’s inequality, we have: n n xrk = k=1 n xq+r k k=1 n yis+1 i=1 g(xk ) ng k=1 n = h(xk ) k(yi ) n nh k=1 n = n nk i=1 n n xk =n· Xnr Xnr = , nr nr−1 xk =n· Xnq+r Xnq+r = q+r−1 , q+r n n k=1 n k=1 n yi =n· i=1 Yns+1 Yns+1 = ns+1 ns Then, we deduce that n k=1 a· r(m+1) (aXnp + bxqk )m+1 xk (cYns − dyks )m ykm = = (anq Xnp+r + bXnq+r )m+1 m(s+1) (cns − d)m Yn (anq Xnp+r + bXnq+r )m+1 (cns m(s+1) d)m Yn · · Xnq+r m+1 Xnp+r + b · q+r−1 r−1 n n dYns+1 m s+1 cYn − ns ms n m(m+1)(q+r−1) , n(m+1)(q+r−1)−ms − and we are done Remark 2.1 If p = q = s = then (2) becomes n k=1 r(m+1) (a + b)m+1 xk (c − d)m ykm r(m+1) (a + b)m+1 Xn (c − d)m Ynm i.e., n (2 ) k=1 r(m+1) xk ykm r(m+1) Xn Ynm n(m+1)(r−1) · n(m+1)(r−1) , D.M Bˇ atinet¸u-Giurgiu, N Stanciu If we consider r = then by (2 ) we obtain n (R) k=1 xm+1 k ykm Xnm+1 , Ynm i.e., just the inequality of J Radon (see, e.g., [2]), with equality if and only if there exists t ∈ R∗+ such that xk = tyk , k = 1, n Remark 2.2 If m = then (2) becomes n (2 ) k=1 (aXnp + bxqk )2 x2r k (cYns − dyks )yk (anq Xnp+r + bXnq+r )2 · 2(q+r−1)−s s+1 s n (cn − d)Yn If we take p = q = s = 0, r = then by (2 ) we obtain n (B) k=1 x2k yk Xn2 Yn But, that is just the inequality of H Bergstrăom A generalization of Nesbitts inequality n k=1 If a ∈ R+ , b, c, d, xk ∈ R∗+ , k = 1, n, Xn = cXnm > d max1 k n xm k , then n (OG) k=1 xk , m ∈ [1, +∞) and (an + b)nm 1−m Xn cnm − d aXn + bxk cXnm − dxm k Proof We have n Un = k=1 n aXn + bxk = cXnm − dxm k n k=1 (aXn + bxk )2 (aXn + bxk )(cXnm − dxm k ) (aXn + bxk )2 = m+1 , m acXnm+1 − adXn xm k + bcXn xk − bdxk k=1 where we apply H Bergstrăom inequality (B) and we deduce that n Un acnXnm+1 − adXn k=1 n k=1 (aXn + bxk ) m xm k + bcXn n n xk − bd k=1 k=1 (anXn + bXn )2 = (acn + bc)Xnm+1 − adXn n k=1 xm k − bd xm+1 k n k=1 xm+1 k Using convexity of the functions f, g : R∗+ → R∗+ , f (x) = xm , g(x) = xm+1 , by Jensen’s inequality we have: n xm k k=1 Xnm nm−1 n xm+1 k and k=1 Xnm+1 , nm Generalizations of some remarkable inequalities and we obtain that (an + b)2 Xn2 adXnm+1 bdXnm+1 (acn + bc)Xnm+1 − − nm−1 nm (an + b)2 nm X 1−m = acnm+1 + bcnm − adn − bd n (an + b)2 nm (an + b)nm 1−m 1−m = X = Xn , n (an + b)(cnm − d) cnm − d which completes the proof Remark 3.1 If a = and b = c = d = m = then we obtain the Nesbitt’s inequality for n variables, i.e., n xk n (N) Xn − xk n−1 Un k=1 If we take n = then by (N) we obtain x1 x2 x3 + + , x2 + x3 x3 + x1 x1 + x2 i.e., Problem 15114, proposed by A.M Nesbitt to Educational Times (1903), 37–38 Remark 3.2 A generalization of (OG) was published in [3], i.e., if a, m ∈ R+ , b, c, d, xk ∈ R∗+ , k = 1, n, Xn = then n (AMM) k=1 n k=1 xk , p ∈ [1, +∞), and cXnm > d max1 aXn + bxk p (cXnm − dxm k ) k n xm k , (an + b)nmp 1−mp X (cnm − d)p n Remark 3.3 A generalization of (AMM) appeared in [4], i.e., if n ∈ N∗ \ {1}, n a ∈ R+ , b, c, d, xk ∈ R∗+ , k = 1, n, Xn = xk , m, p, r, s ∈ [1, +∞), such that k=1 cXnm > d max1 k n xm k , then n k=1 (aXnr + bxrk )s p (cXnm − dxm k ) (anr + b)s mp−rs+1 rs−mp n Xn (cnm − d)p REFERENCES [1] D.M Bˇ atinet¸u-Giurgiu, N Stanciu, Inegalitˇ a¸ti geometrice n poligoane convexe, de tip Bergstră om-Mitrinovi c, Recreatáii Matematice, (2011), 112–115 [2] D.M Bˇ atinet¸u-Giurgiu, Aplicat¸ii la inegalitatea lui J Radon, Gazeta Matematicˇ a – seria B, 7-8-9 (2010), 359–362 [3] D.M Bˇ atinet¸u-Giurgiu, N Stanciu, Problem 11634, The American Mathematical Monthly, 119 (March 2012), 248 [4] D.M Bˇ atinet¸u-Giurgiu, N Stanciu, Nuevas generalizaciones y aplicaciones de la desigualdad de Nesbitt, Revista Escolar de la Olimpiada Iberoamericana de Matematica 47 (nov 2012– feb 2013) “Matei Basarab” National College, Bucharest, Romania “George Emil Palade” Secondary School, Buzˇ au, Romania E-mail: stanciuneculai@yahoo.com