About Surányi s Inequality

Transcription

1 About Suráyi s Iequality Mihály Becze Str Harmaului Sacele Jud Brasov Romaia Abstract I the Milós Schweitzer Mathematical Competitio Hugary Proessor Jáos Suráyi proposed the ollowig problem which is iterestig ad presets a aspect o a theorem I this paper we preset a ew demostratio some iterestig applicatios ad a geeralizatio Theorem 1 Jáos Suráyi I x > 0 = 1 2 the the ollowig iequality holds: 1 x + x x x 1 Proo Usig mathematical iductio or = 2 we obtai x x x 1 x 2 x 1 + x 2 2 which is true We suppose that is true or ad we prove or + 1 Because the iequality is symmetric ad homogeeous we ca suppose that x 1 x 2 x +1 ad x 1 +x 2 ++x = 1 so we must prove the ollowig iequality: x which ca be writte i the orm x +1 +x x +1 x +x +1 From the iductive coditio holds x +1 x x +1 x It remais to prove that: x +1 x x +1 x x x x 1 + x +1 x + x +1 0 x 1 1 x +1 + x +1 x 1 x x + 1 x +1 x but this iequality ca be decomposed i two iequalities i the ollowig maer: First rom the Chebyshev iequality we have: x x 1 0 Secod because x x1 2x = 1 2 1

2 2 the ater additio we have: x + 1 x +1 x 1 +1 or = x x +1 + x x +1 x 1 +1 x +1 + x 1 +1 x +1 x x x +1 x 1 +1 = 0 x 1 x but rom x +1 1 holds the desired iequality I i Theorem 1 we tae = 3 the we obtai: Applicatio 1 I x 1 x 2 x 3 0 the x 1 x x x x 1 x 2 x 3 x 2 1 x 2 + x 3 + x 2 2x 3 + x 1 + x 2 3 x 1 + x 2 which is the well ow Schur s iequality Thereore the iequality o Suráyi has geeralized the Schur iequality Applicatio 2 I a b c deote the sides o triagle ABC s the semiperimeter R the radius o the circumcircle r the radius o the icircle the: 1 R 2r the iequality o Euler 2 s 2 r Rr 3 4R + r 3 s 2 16R 5r Proo I Applicatio 1 we tae: 1 x 1 = a x 2 = b x 3 = c 2 x 1 = s a x 2 = s b x 3 = s c 3 x 1 = r a x 2 = r b x 3 = r c where r a r b r c are the radii o exiscribed circles I I Theorem 1 we tae = 4 the we obtai the ollowig: Applicatio 3 I x 1 x 2 x 3 x 4 0 the x x 1 i<j 4 x i x j x 2 i + x 2 j Remar Because x 2 i + x2 j 2x ix j the 4 x x 1 i<j 4 x 2 i x 2 j but this is the Turevici iequality Thereore the iequality o Suráyi gives a reiemet ad a geeralizatio o Turevici s iequality Applicatio 4 Deote r a r b r c r d ad h a h b h c h d the radii o exiscribed spheres ad the altitudes i tetrahedro ABCD the

3 h a ha r h 3 a 3 1 ra ra r ra 3 where r is the radius o iscribed sphere Proo I Applicatio 3 we tae: 1 x 1 = 1 h a x 2 = 1 h b x 3 = 1 h c x 4 = 1 h d ad 1 h a = 1 r 2 x 1 = 1 r a x 2 = 1 r b x 3 = 1 r c x 4 = 1 r d ad 1 r a = 2 r The iequality o Turevici ca be geeralized i ollowig way: Theorem 2 I x > 0 = 1 2 the x i x j 2 + x 2 1 i<j x 2 Fially we geeralize the iequality o Suráyi i ollowig way: Theorem 3 I a I I R = 1 2 : I R ad ad are covex uctios the: 1 1 ai + a j 1 a + a ij=1 Proo We suppose that a 1 a 2 a so the desired iequality ca be discomposed i the ollowig two iequalities: a + ad 2 a + a a 1 i<j 1 ai + a j 1 a1 + a a 1 a + 2 a + 1 a + a a + 1 ai + a j 1 i<j The iequality 1 is the cosequece o iequalities a + a a 1 a + j=+1 1 a + a j where {1 2 1} but this holds rom Karamata s iequality usig or a a a a + a a

4 4 ad 1 a + a +1 1 a + a +2 1 a + a The iequality o Karamata says that: I : I R is covex x 1 x 2 x ad y 1 y 2 y x 1 y 1 x 1 + x 2 y 1 + y 2 x 1 + x x 1 y 1 + y y 1 x 1 + x x = y 1 + y y the I our case x 1 + x x y 1 + y y x 1 x 2 x = a a a a + a a ad 1 a + a +1 y 1 y 2 y = 1 a + a +2 1 a + a Now we prove the iequality 2 Deote 1 a1 + a a F a 1 a 2 a = i 1 a i + 2 a + 1 iai + a i a 1 ai + a j or which we prove that: F a 1 a 2 a F a 2 a 2 a 3 a 1 i<j F a 1 a 1 a 1 a F a a a = 0 I F a a a a +1 a +2 a cotai a the ollowig expressio i 1 a a + a a a + a a + 1 a + a 1 ai + a 1 i<j j=1 i=+1 a + a a 1 ai + a = i=+1 Deote G a = F a a a a +1 a +2 a where a [a +1 a ] the G a + a = a a 1 1 ai + a 0 i=+1

5 5 because or a + a a 1 a 1 i=+1 i=+1 a i i=+1 1 a i + a which is true Sicee is covex the is icreasig but is covex so a + a a 1 1 a i + a i=+1 which ollows rom Jese s iequality Thereore G is icreasig ad 1 ai + a F a a a a +1 a +2 a F a +1 a +1 a +1 a +2 a which proves the airmatio Remar I i Theorem 3 we tae a = e a ad e a = x = 1 2 the we obtai the iequality o Suráyi Applicatio 5 I a > 0 = 1 2 ad α 2 the α 1 a α 1 α 1 ai + a j + a ij=1 Proo I Theorem 3 we tae a = a α Reereces [1] Mihály Becze: Iequalities mauscript 1982 [2] DS Mitriović JE Pečarić AM Fi: Classical ad New Iequalities i Aalysis Kluwer Academic Publishers 1993 [3] Octogo Mathematical Magazie