Sharpness of the Finsler-Hadwiger inequality

Sharpness of the Finsler-Hadwiger inequality Cezar Lupu Departmen t of Mathematics-Informatics Univ ersit y of Buc harest Buc harest, Ro mania R O-01001 4 lupucezar@y ahoo.com Cosmin P ohoat ¸ ˘ a T udor Vianu National College Buc harest, Ro mania R O-01001 4 pohoata_cos min2000@yahoo. com de dic ate d to the m emory o f the gr e at pr ofes s or, Alexandru Lup a ¸ s 1 In tro d uction & Preliminarie s The Hadwiger-Finsler inequalit y is kno wn in literature o f mat hematics as a gener- alization of the following Theorem 1.1 In any triangle AB C with the side lenghts a, b, c and S its ar e a, the fol lowing ine quality is v alid a 2 + b 2 + c 2 ≥ 4 S √ 3 . This inequalit y is due to W eitzen b o c k, Math. Z, 137- 146, 1 9 19, but this has also app eared at In ternational Mathematical Olympiad in 1961. In [7.], one can find elev en pro ofs. In fact, in an y triangle AB C the follo wing seque nce of inequalities is v alid: a 2 + b 2 + c 2 ≥ ab + bc + ca ≥ a √ bc + b √ ca + c √ ab ≥ 3 3 √ a 2 b 2 c 2 ≥ 4 S √ 3 . A stronger v ersion is t he one found b y Finsler and Hadwiger in 1938, whic h states that ([2.]) Theorem 1.2 In any triangle AB C with the side lenghts a, b, c and S its ar e a, the fol lowing ine quality is v alid a 2 + b 2 + c 2 ≥ 4 S √ 3 + ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 . 1 In [8.] the first author of this note g a v e a simple pro of only by using AM-GM and the follow ing inequality due to Mitrino vic: Theorem 1.3 In any triangle AB C with the side lenghts a, b, c and s its semip erime- ter and R its cir cumr adius, the fol lowing i n e quality holds s ≤ 3 √ 3 2 R. This inequalit y also app ear s in [3.]. A nice inequalit y , sharp er than Mitrino vic and equiv alent to the first theorem is the follow ing: Theorem 1.4 In an y triangle AB C with sides of lenghts a, b, c and with inr a d ius of r , cir cumr a dius of R a n d s its semip erimeter the fol lowing ine quality holds 4 R + r ≥ s √ 3 . In [4.], W u gav e a nice sharpness and a generalization of the F insler-Hadwiger inequalit y . No w, w e give an algebraic inequalit y due t o I. Sc h ur ([5.]), namely Theorem 1.5 F or a ny p ositive r e al numb ers x, y , z and t ∈ R the fol lowing in e qual- ity holds x t ( x − y )( x − z ) + y t ( y − x )( y − z ) + z t ( z − y ) ( z − x ) ≥ 0 . The most common case is t = 1, whic h has the f ollo wing equiv alen t form: x 3 + y 3 + z 3 + 3 xy z ≥ xy ( x + y ) + y z ( y + z ) + z x ( z + x ) whic h is equiv alen t to x 3 + y 3 + z 3 + 6 xy z ≥ ( x + y + z )( xy + y z + z x ) . No w, using the iden tity x 3 + y 3 + z 3 − 3 xy z = ( x + y + z )( x 2 + y 2 + z 2 − xy − y z − z x ) one can easily deduce that 2( xy + y z + z x ) − ( x 2 + y 2 + z 2 ) ≤ 9 xy z x + y + z . ( ∗ ) Another in teresting case is t = 2. W e hav e 2 x 4 + y 4 + z 4 + xy z ( x + y + z ) ≥ xy ( x 2 + y 2 ) + y z ( y 2 + z 2 ) + z x ( z 2 + x 2 ) whic h is equiv alen t to x 4 + y 4 + z 4 + 2 xy z ( x + y + z ) ≥ ( x 2 + y 2 + z 2 )( xy + y z + z x ) . ( ∗∗ ) No w, let’s rewrite theorem 1.2. a s 2( ab + bc + ca ) − ( a 2 + b 2 + c 2 ) ≥ 4 S √ 3 . ( ∗ ∗ ∗ ) By squaring ( ∗ ∗ ∗ ) and using Heron form ula w e obtain 4 X cy c ab ! 2 + X cy c a 2 ! 2 − 4 X cy c ab ! X cy c a 2 ! ≥ 3( a + b + c ) Y ( b + c − a ) whic h is equiv alen t to 6 X cy c a 2 b 2 + 4 X cy c a 2 bc + X cy c a 4 − 4 X cy c ab ( a + b ) ≥ 3( a + b + c ) Y ( b + c − a ) . By making some elemen tary calculations w e get 6 X cy c a 2 b 2 +4 X cy c a 2 bc + X cy c a 4 − 4 X cy c ab ( a + b ) ≥ 3( a + b + c ) X cy c ab ( a + b ) − X cy c a 3 − 2 abc ! . W e obtain the equiv alent inequalities X cy c a 4 + X cy c a 2 bc ≥ X cy c ab ( a 2 + b 2 ) a 2 ( a − b )( a − c ) + b 2 ( b − a )( b − c ) + c 2 ( c − a )( c − b ) ≥ 0 , whic h is nothing else than Sc h ur’s inequalit y in the pa rticular case t = 2. In what follo ws w e will give another form of Sc h ur’s inequality . That is Theorem 1.6 F or any p ositive r e als m, n, p , the fol lowing in e quality holds mn p + np m + pm n + 9 mnp mn + np + pm ≥ 2( m + n + p ) . Pr o of. W e denote x = 1 m , y = 1 n and z = 1 p . W e obtain the equiv alen t inequalit y x y z + y z x + z xy + 9 x + y + z ≥ 2( xy + y z + z x ) xy z ⇔ 2( xy + y z + z x ) − ( x 2 + y 2 + z 2 ) ≤ 9 xy z x + y + z , whic h is ( ∗ ). 3 2 Main re sults In the previous section w e stated a sequence of inequalities stronger than W eitzen- b o ck inequalit y . In fact, one can prov e that the follo wing sequence of inequalities holds a 2 + b 2 + c 2 ≥ ab + bc + ca ≥ a √ bc + b √ ca + c √ ab ≥ 3 3 √ a 2 b 2 c 2 ≥ 18 R r , where R is the c ircumradius and r is the inradius of the triangle with sides of le ngh ts a, b, c . In this momen t, one exp ects to ha v e a strong er Finsler-Hadwiger inequalit y with 18 Rr instead of 4 S √ 3. Unfo rtunately , the following inequalit y holds true a 2 + b 2 + c 2 ≤ 18 R r + ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 , b ecause it is equiv alen t to 2( ab + bc + ca ) − ( a 2 + b 2 + c 2 ) ≤ 18 Rr = 9 abc a + b + c , whic h is ( ∗ ) again. Now , we are ready to prov e the first refinemen t of the Finsler- Hadwiger inequalit y: Theorem 2.1 In an y triangle AB C with the side lenghts a, b, c with S its ar e a, R the cir cumr adius and r the inr adius of the triangle AB C the fol lowing ine q uali ty is valid a 2 + b 2 + c 2 ≥ 2 S √ 3 + 2 r (4 R + r ) + ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 . Pr o of. W e rewrite the inequality as 2( ab + bc + ca ) − ( a 2 + b 2 + c 2 ) ≥ 2 S √ 3 + 2 r (4 R + r ) . Since, ab + bc + ca = s 2 + r 2 + 4 Rr , it follows immediately that a 2 + b 2 + c 2 = 2( s 2 − r 2 − 4 R r ). The inequality is equiv alent to 16 Rr + 4 r 2 ≥ 2 S √ 3 + 2 r (4 R + r ) . W e finally obtain 4 R + r ≥ s √ 3 , whic h is exactly theorem 1.4 .  The second refinemen t o f the Finsler-Hadwiger inequalit y is the f ollo wing 4 Theorem 2.2 In an y triangle AB C with the side lenghts a, b, c with S its ar e a, R the cir cumr adius and r the inr adius of the triangle AB C the fol lowing ine q uali ty is valid a 2 + b 2 + c 2 ≥ 4 S r 3 + 4( R − 2 r ) 4 R + r + ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 . Pr o of. In theorem 1.6 w e put m = 1 2 ( b + c − a ) , n = 1 2 ( c + a − b ) a nd p = 1 2 ( a + b − c ). W e get X cy c ( b + c − a )( c + a − b ) ( a + b − c ) + 9( b + c − a )( c + a − b )( a + b − c ) X cy c ( b + c − a )( c + a − b ) ≥ 2( a + b + c ) . Since ab + bc + ca = s 2 + r 2 + 4 Rr (1) and a 2 + b 2 + c 2 = 2( s 2 − r 2 − 4 Rr ) (2), we deduce X cy c ( b + c − a )( c + a − b ) = 4 r (4 R + r ) . On the other hand, b y Heron’s form ula we ha ve ( b + c − a )( c + a − b )( a + b − c ) = 8 sr 2 , so our inequalit y is equiv alent to X cy c ( b + c − a )( c + a − b ) ( a + b − c ) + 18 sr 4 R + r ≥ 4 s ⇔ X cy c ( s − a )( s − b ) ( s − c ) + 9 sr 4 R + r ≥ 2 s ⇔ X cy c ( s − a ) 2 ( s − b ) 2 + 9 s 2 r 3 4 R + r ≥ 2 s 2 r 2 . No w, according to the identit y X cy c ( s − a ) 2 ( s − b ) 2 = X cy c ( s − a )( s − b ) ! 2 − 2 s 2 r 2 , w e ha v e X cy c ( s − a )( s − b ) ! 2 − 2 s 2 r 2 + 9 s 2 r 3 4 R + r ≥ 2 s 2 r 2 . And since X cy c ( s − a )( s − b ) = r (4 R + r ) , it follo ws that r 2 (4 R + r ) 2 + 9 s 2 r 3 4 R + r ≥ 4 s 2 r 2 , 5 whic h rewrites as  4 R + r s  2 + 9 r 4 R + r ≥ 4 . F rom the iden tities mentioned in (1) and (2) we deduce that 4 R + r s = 2( ab + bc + ca ) − ( a 2 + b 2 + c 2 ) 4 S . The inequalit y rewrites a s  2( ab + bc + ca ) − ( a 2 + b 2 + c 2 ) 4 S  2 ≥ 4 − 9 r 4 R + r ⇔  ( a 2 + b 2 + c 2 ) − (( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 ) 4 S  2 ≥ 3 + 4( R − 2 r ) 4 R + r ⇔ a 2 + b 2 + c 2 ≥ 4 S r 3 + 4( R − 2 r ) 4 R + r + ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 .  Remark. F rom Euler inequality , R ≥ 2 r , w e obtain theorem 1.2. 3 Applicatio ns In this section w e illustrate some basic applications of the second refinemen t of Finsler-Hadwiger inequalit y . W e b egin with Application 1. In any triangle AB C with the sides of lenghts a, b, c the fol low- ing ine quality holds 1 b + c − a + 1 c + a − b + 1 c + a − b ≥ 1 2 r r 4 − 9 r 4 R + r . Solution. F rom ( b + c − a )( c + a − b )( a + b − c ) = 4 r (4 R + r ) , it is quite easy to o bserv e that 1 b + c − a + 1 c + a − b + 1 a + b − c = 4 R + r 2 sr . No w, applying the inequalit y  4 R + r s  2 + 9 r 4 R + r ≥ 4 , 6 w e get  1 b + c − a + 1 c + a − b + 1 a + b − c  2 = 1 4 r 2  4 R + r s  2 ≥ 1 4 r 2  4 − 9 r 4 R + r  . The giv en inequalit y follows immediately .  Application 2. In any triangle AB C with the sides of lenghts a, b, c the fol low- ing ine quality holds 1 a ( b + c − a ) + 1 b ( c + a − b ) + 1 c ( a + b − c ) ≥ r 8 R  5 − 9 r 4 R + r  . Solution. F rom the following iden tit y X cy c ( s − a )( s − b ) c = r ( s 2 + (4 R + r ) 2 ) 4 sR = S 4 R 1 +  4 R + r p  2 ! . Using the inequalit y  4 R + r s  2 + 9 r 4 R + r ≥ 4 , w e ha v e X cy c ( s − a )( s − b ) c ≥ S 4 R  5 − 9 r 4 R + r  . In this momen t, the problem follo ws easily .  Application 3. In any triangle AB C with the sides of lenghts a, b, c the fol low- ing ine quality holds 1 ( b + c − a ) 2 + 1 ( c + a − b ) 2 + 1 ( a + b − c ) 2 ≥ 1 r 2  1 2 − 9 r 4(4 R + r )  . Solution. F rom ( b + c − a )( c + a − b )( a + b − c ) = 4 r (4 R + r ) , it follows that ( b + c − a ) 2 + ( c + a − b ) 2 + ( a + b − c ) 2 = 4( s 2 − 2 r 2 − 8 Rr ) and ( b + c − a ) 2 ( c + a − b ) 2 +( a + b − c ) 2 ( c + a − b ) 2 +( b + c − a ) 2 ( a + b − c ) 2 = 4 r 2  (4 R + r ) 2 − 2 s 2  . W e get 1 ( b + c − a ) 2 + 1 ( c + a − b ) 2 + 1 ( a + b − c ) 2 = 1 4  (4 R + r ) 2 s 2 r 2 − 2 r 2  . 7 No w, applying the inequalit y  4 R + r s  2 + 9 r 4 R + r ≥ 4 , w e ha v e 1 ( b + c − a ) 2 + 1 ( c + a − b ) 2 + 1 ( a + b − c ) 2 ≥ 1 4 r 2  2 − 9 r 4 R + r  = 1 r 2  1 2 − 9 r 4(4 R + r )  .  Application 4. In any triangle AB C with the sides of lenghts a, b, c the fol low- ing ine quality holds a 2 b + c − a + b 2 c + a − b + c 2 a + b − c ≥ 3 R r 4 − 9 r 4 R + r . Solution. Without loss of generality , w e assume that a ≤ b ≤ c . It follows quite easily that a 2 ≤ b 2 ≤ c 2 and 1 b + c − a ≤ 1 c + a − b ≤ 1 a + b − c . Applying Cheb y- shev’s inequalit y , we ha v e a 2 b + c − a + b 2 c + a − b + c 2 a + b − c ≥  a 2 + b 2 + c 2 3   1 b + c − a + 1 c + a − b + 1 c + a − b  . No w, the first application and the inequality a 2 + b 2 + c 2 ≥ 18 R r solve s the problem.  Application 5. I n any triangle AB C with the sides of lenghts a, b, c and with the exr adii r a , r b , r c c orr esp ondin g to the triangle AB C , the fol lowing ine quality holds a r a + b r b + c r c ≥ 2 r 3 + 4( R − 2 r ) 4 R + r . Solution. F rom the w ell-kno wn relations r a = S s − a and the analogues, the in- equalit y is equiv alen t to a r a + b r b + c r c = 2( ab + bc + ca ) − ( a 2 + b 2 + c 2 ) 2 S ≥ 2 S r 3 + 4( R − 2 r ) 4 R + r . The last inequalit y follo ws from theorem 2.2 immediately .  Application 6. I n any triangle AB C with the sides of lenghts a, b, c and with the exr adii r a , r b , r c c orr esp ondin g to the triang l e AB C and with h a , h b , h c b e the altitudes of the triangle AB C , the fol lowing ine quality holds 8 1 h a r a + 1 h b r b + 1 h c r c ≥ 1 S r 3 + 4( R − 2 r ) 4 R + r . Solution. F rom t he we ll-kno wn r elat io ns in tria ngle AB C , h a = 2 S a , r a = S s − a w e ha v e 1 h a r a = a ( s − a ) 2 S 2 . Doing the same thing for the analogues and a dding them up w e get 1 h a r a + 1 h b r b + 1 h c r c = 1 2 S 2 ( a ( s − a ) + b ( s − b ) + c ( s − c )) . On the other hand by using theorem 2.2 in the fo rm a ( s − a ) + b ( s − b ) + c ( s − c ) ≥ 2 S r 3 + 4( R − 2 r ) 4 R + r w e obtain the desired inequality .  Application 7. In any triangle AB C with the sides of lenghts a, b, c the fol low- ing ine quality holds true tan A 2 + tan B 2 + tan C 2 ≥ r 3 + 4( R − 2 r ) 4 R + r . Solution. F rom the cosine law we get a 2 = b 2 + c 2 − 2 bc cos A . Since S = 1 2 bc sin A it follo ws that a 2 = ( b − c ) 2 + 4 S · 1 − cos A sin A . On the other hand by the trigonometric form ulae 1 − cos A = 2 sin 2 A 2 and sin A = 2 sin A 2 cos A 2 w e get a 2 = ( b − c ) 2 + 4 S tan A 2 . Doing the same for all sides of the triangle AB C and adding up w e obtain a 2 + b 2 + c 2 = ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 + 4 S  tan A 2 + tan B 2 + tan C 2  . No w, b y theorem 2.2 t he inequalit y follo ws.  Application 8. I n any triangle AB C with the sides of lenghts a, b, c and with the exr adii r a , r b , r c c orr esp ondin g to the triangle AB C , the fol lowing ine quality holds r a a + r b b + r c c ≥ s (5 R − r ) R (4 R + r ) . 9 Solution. It is w ell- kno wn that the following identit y is v alid in an y triangle ABC r a a + r b b + r c c = (4 R + r ) 2 + s 2 4 Rs . So, the inequalit y r ewrites as (4 R + r ) 2 s 2 + 1 ≥ 4(5 R − r ) 4 R + r , whic h is equiv alen t with  4 R + r s  2 + 9 r 4 R + r ≥ 4 .  Ac kno wledgmen t. The authors would like to thank to Nicola e Constan tinescu , from U niv ersit y of Craio v a a nd to Marius G hergu, from the Institute of Mathematics of the Romanian Academ y for useful suggestions. This pap er has b een completed while the first autho r participated in the summer sc ho ol on Critic al Point the o ry a nd its applic ations organized in Cluj-Nap o ca city . W e are kindly grateful to pro fessors Vicent ¸ iu R ˘ adulescu fr o m the Institute o f Mathematics of the Romanian Academ y and to Csaba V arg a f rom Bab e¸ s-Boly ai Univ ersit y , Cluj-Nap o ca. References [1] Roland W eitzen b o ck, Ub er eine Ungleic h ung in der Dreieck sgeometrie, Math.Z , ? , (1919) , 137-146. [2] P . F insler, H. Hadwiger, Einige Relationen im Dreic k , Comment Math. Helv. , 10 , (1938) , 316-3 26. [3] O.Bottema, R.Z. Djo rdjevic, R.R . Janic, D.S. Mitrinovic, P .M. V asic , Geomet- ric inequalities, Wolters-No or dhoff , Gr oningen (1969). [4] Shank e W u, Generalization and Sharpness of Finsler-Hadwiger’s inequalit y and its applications, Mathematic al Ine qualities and Applic ations , 9 , no. 3, (20 0 6), 421-426 . [5] G.N. W atson, Sc h ur’s inequalit y , The Mathema tic al Gazzette , 39 , (1 955), 20 7 - 208. 10 [6] John Steinig, Inequalities concerning the inradius and circumradius of a trian- gle, Elemente der Mathematik , 18 , (1963), 12 7 -131. [7] Arth ur Engel, Problem solving strategies, Springer V erlag (199 8). [8] Cezar Lupu, An elemen tary pro of of the Hadwiger-F insler inequalit y , A rhime de Magazine, 3 , no.9- 1 0, (2003), 18-19. 11