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Võ Quốc Bá Cẩn, Phạm Thị Hằng:The sums of square technique
Võ Quc Bá Cn Phm Th Hng Copyright © Vo Quoc Ba Can 1 THE SUMS OF SQUARE TECHNIQUE I. Theorem. Consider the following inequality 422332 ( )0 cyc cyc cyc cyc cyc m a n a b p a b g ab m n p g a bc +++−+++≥ ∑∑∑∑∑ With ,, abc be real numbers. Then this inequality holds when 22 0 3() m mmn p pgg > +≥++ . Proof. We rewrite the inequality as 422 222 3 2 32 () 0 cyc cyc cyc cyc cyc cyc cyc cyc m a ab m n ab abc p ab abc g ab a bc −++ −+ − + −≥ ∑∑ ∑ ∑ ∑∑ ∑∑ Note that 4 2 2 2 22 3 2 3 2 22 22 22 22 3 2 3 2 22 22 1 () 2 () 11 ()( )()()( 2) 33 () () cyc cyc cyc cyc cyc cyc cyc cyc cyc cyc cyc cyc cyc cyc cyc cyc cy a ab ab ab abc bc abc bca b bc a b ab bc ca a b a b ab ac bc ab a bc ca ab c ca a b caab −=− −=−= − =− − + ++ − = − +− −=−= − =− ∑∑∑ ∑∑∑∑∑ ∑ ∑∑ ∑∑∑∑∑ 22 22 11 ( ) ( ) ( )( 2) 33 c cyc cyc ab bc ca a b a b ab bc ca − ++ − =− − +− ∑ ∑∑ Then the inequality is equivalent to 222 22 222 1 ( ) ( )[()(2)(2)] 23 ()0 cyc cyc cyc cyc m a b a b p g ab p g bc p g ca m n a b a bc − + − − − + ++ ++ −≥ ∑∑ ∑∑ Moreover 2222 22 1 [()(2)(2)] 6() cyc cyc cyc a b a bc p g ab p g bc p g ca p pgg − = − − + ++ ++ ∑∑∑ The inequality becomes Võ Quc Bá Cn Phm Th Hng Copyright © Vo Quoc Ba Can 2 222 22 2 22 222 1 ( ) ( )[()(2)(2)] 23 [( ) (2 ) (2)]0 6() 1 [3( )()(2)(2)] 18 cyc cyc cyc cyc m a b a b p g ab p g bc p g ca mn p g ab p g bc p g ca p pgg ma b p gab p gbc p gca m − + − − − + ++ + + −−+++≥ ++ ⇔ − +− − + ++ ∑∑ ∑ ∑ 22 2 22 3() [()(2)(2)]0 18() cyc mmn p pgg p g ab p g bc p g ca mp pgg +−−− + −−+++≥ ++ ∑ From now, we can easily check that if 22 0 3() m mmn p pgg > +≥++ then the inequality is true. Our theorem is proved. J II. Application. Example 1 . (Vasile Cirtoaje) Prove that 2222 333 ( ) 3( ). a b c ab bc ca ++ ≥ ++ Solution. The inequality is equivalent to 4 223 20 cyc cyc cyc a ab ab + −≥ ∑∑∑ From this, we get 1,2, 3,0 mnpg == =−= , we have 22 22 10 3 ( ) 3 1 (1 2) ( 3) ( 3) 0 0 0 m mmn p pgg => + − − − =⋅⋅ + −− −−⋅− = Then using our theorem, the inequality is proved. J Example 2 . (Võ Quc Bá Cn) Prove that ( ) 444 333 3 1 ( ) 3( ). a b c abca b c ab bc ca +++ − ++≥ ++ Solution. We have 1,0, 3,0 mnpg ===−= and ( ) ( ) 2 222 10 3( ) 31(10) 3 3000 m mmn p pgg => + − − − = ⋅⋅ + −− −− ⋅ − = Then the inequality is proved. J Example 3 . (Phm Vn Thun) Prove that 444 333 7( ) 10( ) 0. abc abbcca +++ ++≥ Solution. We will prove the stronger result, that is 4 43 17 7 10 27 cyc cyc cyc aaba +≥ ∑∑∑ Võ Quc Bá Cn Phm Th Hng Copyright © Vo Quoc Ba Can 3 4223 32 86 51 101 34 102 0 86 51 101 34 cyc cyc cyc cyc cyc a a b a b ab a bc m n p g ⇔−+−−≥ = =− ⇒ = =− ∑∑ ∑∑∑ Moreover 22 22 860 3 ( ) 3 86 (86 51) 101 101 ( 34) ( 34) 1107 0 m mmn p pgg => +−−−=⋅⋅−− − ⋅−−− = > Then the inequality is proved. J Example 4 . (Vình Quý) Let , , 0, 1. a b c abc >= Prove that 2 22 111 3. 111aa bb cc ++≤ −+ −+ −+ Solution. On Mathlinks inequality forum, I posted the following proof: Lemma. If , , 0, 1, a b c abc >= then 222 111 1. 111aa bb cc ++≥ ++ ++ ++ Proof. From the given condition ,,0,1 a b c abc >= , there exist ,,0 xyz > such that 2 2 2 . yz a x zx b y xy c z = = = And then, the inequality becomes 4 4 2 22 1 cyc x x xyz yz ≥ ++ ∑ By the Cauchy Schwarz Inequality, we get 2 22 2 22 4 4 2 22 4 2 22 4 22 2 4 22 1 ()2 cyc cyc cyc cyc cyc cyc cyc cyc cyc cyc x xx x x xyz yz x xyz yz x yz xyz x yz ≥ = ≥= ++ +++++ ∑ ∑∑ ∑ ∑ ∑∑ ∑ ∑∑ Our lemma is proved. Now, using our lemma with note that 222 222 111 ,,0 , 111 1 abc abc > ⋅⋅= we get Võ Quc Bá Cn Phm Th Hng Copyright © Vo Quoc Ba Can 4 422 42 42 42 22 22 22 1 2( 1) 124 111 ( 1) ( 1) 1 1 44 ( 1)( 1) 1 1 cyc cyc cyc cyc cyc cyc xxx xx xx xx xx xx xx xx xx xx ++ ≥⇔≤⇔≤ ++ ++ ++ ++ + −+ ⇔ ≤⇔+≤ ++ −+ −+ ++ ∑∑∑ ∑ ∑∑ Using our lemma again, we can get the result. J Now, I will present another proof of mine based on this theorem Since , , 0, 1, a b c abc >= there exists ,,0 xyz > such that ,, yzx abc xyz === then our inequality becomes 22 22 22 22 22 22 3 3 ( 2) 39433 cyc cyc cyc cyc x x x xy x xyy x xyy x xyy x xyy − ≤⇔≤⇔−≥⇔≥ −+ −+ −+ −+ ∑∑∑∑ By the Cauchy Schwarz Inequality, we get 2 2 2222 22 ( 2) ( 2)( ) ( 2) cyc cyc cyc xy xyxxyy xy x xyy − − −+≥− −+ ∑∑∑ It suffices to show that 2 2 222 4 22 3 32 ( 2) 3 ( 2)( ) 10 39 25 16 8 0 cyc cyc cyc cyc cyc cyc cyc x y x yx xyy x x y x y xy x yz − ≥ − −+ ⇔+ −−−≥ ∑∑ ∑∑ ∑ ∑∑ From this, we get 10, 39, 25, 16 mnpg = = =−− and 22 22 100 3 ( ) 3 10 (10 39) ( 25) ( 25) ( 16) ( 16) 189 0 m mmn p pgg => + − − − = ⋅ ⋅ + −− −− ⋅− −− = > Then using our theorem, the inequality is proved. J III. Some problems for own study. Problem 1 . (Vasile Cirtoaje) Prove that 444333 333 2( ). a b c ab bc ca ab bc ca +++++≥ ++ Problem 2 . (Phm Vn Thun, Võ Quc Bá Cn) Prove that 3334 8 ( ) ( ) ( ) ( ). 27 aab bbc cca abc + + + + + ≥ ++ Problem 3 . (Phm Kim Hùng) Prove that 444 2 333 1 ( ) 2( ). 3 a b c ab bc ca a b b c c a +++ ++ ≥ + + Võ Quc Bá Cn Student Can Tho University of Medicine and Pharmacy, Can Tho, Vietnam E-mail: can_hang2007@yahoo.com Võ Quc Bá Cn Phm Th Hng Copyright © Vo Quoc Ba Can 5 . 1 THE SUMS OF SQUARE TECHNIQUE I. Theorem. Consider the following inequality 422332 ( )0 cyc cyc cyc cyc cyc m a n a b p a b g ab m n p g a bc +++−+++≥ ∑∑∑∑∑ With ,, abc be real numbers. Then. we can get the result. J Now, I will present another proof of mine based on this theorem Since , , 0, 1, a b c abc >= there exists ,,0 xyz > such that ,, yzx abc xyz === then our inequality becomes 22. I posted the following proof: Lemma. If , , 0, 1, a b c abc >= then 222 111 1. 111aa bb cc ++≥ ++ ++ ++ Proof. From the given condition ,,0,1 a b c abc >= , there exist ,,0 xyz > such