AN WEIGHTED POWER LESSELS-PELLING INEQUALITY

Transcription

1 AN WEIGHTED POWER LESSELS-PELLING INEQUALITY MIHÁLY BENCZE AND MARIUS DRĂGAN Absrac. Lessels-Pelling inequaliy saes ha in a riangle he sum of wo angle bisecs and a medians is less equal han he produc beween 3 and semiperimeer. The purpose of his aricle is o find a powered opion of his inequaliy wih some supplemenary condiion, f he weighs and some chains of inequaliies which represen refinemens of powered Lessels-Pelling inequaliy.. Inroducion Le ABC be a riangle. We shall denoe a = BC, b = CA, c = AB, s = a + b + c he semiperimeer, m a he medians from A, w b, w c he bisecs from B and C. In 97 J. Garfunkel [] conjuncured on basis of compuer check he following inequaliy m a + w b + h c s 3, where h c represen he aliude from C. In 976 C.S. Gardner [] prove by mean of sequence of elemenary ransfmaion and differeniable of a funcion. In 977 G.S. Lessels and M.J. Pelling give in [] sronger inequaliy m a + w b + w c s 3 using he compuer check. In 980 B.E. Pauwo, R.S.D. Thomas and Chung-Lie Wang give in [3] a proof of his inequaliy. In [] appear a new proof of C. Tănăsescu. In 98 L. Panaiopol [5] find an elemenary soluion of Lessels-Pelling inequaliy. În [6] and [7] M. Drăgan give a simple proof and some refinemens of Lessels-Pelling inequaliy.. Main resuls In he following we use he following noaion x = a s, y = b s, z = c s, u = y, v = z, s = u + v, p = u v, y E = + z ) x, s = βu + γv, = 3α α +,.) where α, β, γ are posiive numbers such ha β γ)b c) 0 and α+β+γ =. 00 Mahemaics Subjec Classificaion.???????? Key wds and phrases.??????????????

2 Mihály Bencze and Marius Drăgan Lemma.. Accding wih above noaion, are rue he inequaliies: i) E ii) E s.) α) s.3) Proof. i) From.) we have + u E = + v u v ) u + v ) [ + u = + v ) p u + v )] u + v ) + s = p ) p s p ) s p ) + s = s + p s ) s + s = s ii) We have s = βu + γv β + γ)u + v) = α)s since β γ)b c) 0, i resuls ha β γ)u v) 0. From.) and.) i resuls.3)..) Theem. The weighed power Lessels-Pelling inequaliy). In every ABC riangle is rue he following inequaliy αm a + βw b + γw c S.5) where α, β, γ are posiive numbers such ha β γ)b c) 0 and α+β+γ =. Proof. Since w b ss b) and w c ss c) o prove.5) i will be sufficien o prove ha αm a + β ss b) + γ ss c) s.6) Accding.) inequaliy,.6) may be wrien as y α + z ) x + β y + γ z αe + S.7) From.) and.) we have ha αe + S α S + α I resuls ha o prove.7), i will be sufficien o prove ha S α S + α S.8)

3 An weighed power Lessels-Pelling inequaliy 3 We noe ha S = y+ z y z) = x = So, we have S <. S α S <. We consider he funcion f : 0, ) R, fs ) = α + α)s wih f S ) = αs + α. We solve he equaion f S S ) = 0 wih α) he roo S 0 = 6α α +. Since f 0) 0, f ) 0, i resuls ha S 0 is a maximum poin f f. So we have [ ] α) α) fs ) fz 0 ) = α 6α + α + 6α α + α = α 6α α + + α) 6α α + 3α = α + = which prove.8). So, we have he following refinemen of.5). Theem.. In every ABC riangle is rue he following chain of inequaliies: αm a + βw b + γw c αm a + β ss b) + γ ss c) s s s b + s c ) α + α ) ss b) + ss c) s where α, β, γ are posiive numbers such ha β γ)b c) 0 and α + β + γ =. Remark.. If we ake in.5) α = β = γ we obain he Lessels-Pelling inequaliy. Theem.3. In every ABC riangle are rue he following inequaliies: i) m a α m a s + α s.9) ii) α + α α s m a s + α s α α) β s b + γ s c).0)

4 Mihály Bencze and Marius Drăgan iii) m a s + α α s α) β s b + γ ) s c α α) β s b + γ s c).) where α, β, γ are posiive numbers such ha α+β+γ = and β γ)b c) 0. Proof. i) From.), inequaliy.9) may be wrien as E α E + α ) E α 0. ii) From.3) we have α E + α α = + α α α α) S + α α α) S.) Taking in.) he noaion from.) we obain he equaliy from saemen. iii) The firs inequaliy is jus inequaliy.3). We will prove he second inequaliy which is equivalen wih [ α) S + α α α α) S ] α) S y α 0. Theem.. In every ABC riangle is rue he following chain of inequaliies: ii) i) αm a + βw b + γw c αm a + β ss b) + γ ss c) ma + α s s + β ss b) + γ ss c) + α s α) β s b + γ ) s c αm a +βw b +γw c α + β ss b) + γ ss c) s s α) β s b + γ + s c) βwb + γw c α) β s b + γ +βwb s c) +γw c +α s + α s α) β s b + γ s c + β ss b) + γ ss c) s, )

5 An weighed power Lessels-Pelling inequaliy 5 where α, β, γ are posiive numbers such ha α + β + γ = and β γ)b c) 0. Proof. i) The firs inequaliy i follows from inequaliies w b ss b) and w c ss c). The second inequaliy i resuls from.9). Also he hird i follows from.0). I remain o prove ha + α s α) β s b + γ ) s c +β ss b) + γ ss c) s..3) Using.), inequaliy.3) may be wrien as + α α) S + S S α) S + α ) α) 0 S α) α) S + 0 S + α) 0. ii) The firs inequaliy from his chain i follows from.). The second i resuls also from.). The hird is rue as w b ss b), w c ss c), and las inequaliy from chain is he inequaliy from i) Collary.. In every ABC riangle is rue he following inequaliy α αm a + βw b + γw c + β + γ) s,.) where α, β, γ are posiive numbers such ha b c)β γ) 0. Proof. If we ake α we obain.). α α + β + γ, β β α + β + γ, γ γ α + β + γ in.5), Collary.. In every ABC riangle is rue he following inequaliy w m a w a + m c w b + m b w c a + m b + m c ). Proof. Since b c)m c m b ) 0, if we ake α = w a, β = m c, γ = m b in.), we obain he inequaliy from he saemen. Collary.3. In every ABC riangle is rue he following inequaliy: m a m b m c + w a w b m c + w a m b w c m b m c + w am b + m c )..5) Proof. Since b c) ) 0, if we ake in.) α =, β =, m b m c w a m b γ = m c, we obain he inequaliy from he saemen.

6 6 Mihály Bencze and Marius Drăgan References [] Garfunkel, J., Problem E 50, Amer. Mah. Monhly 8 97), soluion by C. S. Garden) Amer. Mah. Monly ), [] Lessels, G.S. and Pelling, M.J., An inequaliy f he sum of wo angle bisecs and a median, Univ. Beograd, Publ. Elecroehn. Fak. Ser. Ma. Fiz. no. 577, no ), 59 6 [3] Pauwo, B.E., Thomas, R.S.D. and Chung-Lie Wang, The riangle inequaliy of Lessels and Pelling, Univ. Beograd, Publ. Elecroehn. Fak. Ser. Ma. Fiz. no. 678, no. 75 [] Mirinović, D.S., Pečarić, J.E. and Volonec, V., Recen Advances in Geomeric inequaliies, -3 [5] Panaiopol, L., Some abou geomeric inequaliies, G.M. B no. 8 98), Romanian) [6] Drăgan, M., Some geomeric inequaliies, Rev. Arhimede no ), 6 8 Romanian) [7] Drăgan M., Some exensions and refinemens of Lessels-Pelling inequaliy, G.M. B. no ), 8 8 Romanian) [8] Ocogon Mahemaical Magazine 997 0) Hărmanului 6, Săcele-Négyfalu, Braşov, Romania address: benczemihaly@yahoo.com 63 Timişoara 35, Bl. 0D6, Sc. E, e. 7, Ap. 76, Sec. 6, Bucureşi, Romania address: marius.dragan005@yahoo.com