Inequality season 1 koopa VẺ ĐẸP CỐT LÕI CỦA BẤT ĐẲNG THỨC

Vẻ đẹp cốt lõi Bất đẳng thức Lời nói đầu: Bất đẳng thức chuyên đề hay khó chương trình Tốn, có mặt thi Tốn đồng thời đòi hỏi ta phải có lượng kiến thức vững vàng để giải BĐT Nhưng quyến rũ, tao nhã mà BĐT thu hút nhiều hệ học sinh – sinh viên nước nhà tham gia giải sáng tạo, phát triển thêm Để trì say mê tư sáng tạo ấy, trang mạng, diễn đàn chia sẻ nhiều topic, tài liệu BĐT bổ ích cho người, rắc rối chưa xếp trọn vẹn Có thể nhiều người nghĩ BĐT học chút cảm thấy nhàm chán, lượng kiến thức rộng “khô khan” việc trình bày lượng kiến thức ấy, đặc biệt phạm vi THPT trở lên, cố gắng tóm gọn trình bày rõ ràng lại cho người thông qua file tài liệu để giúp người nhận vẻ đẹp hoàn mĩ ẩn chứa BĐT ứng dụng File tài liệu soạn dựa vào tài liệu nhiều giáo viên, giáo sư tiếng toàn giới Trần Quốc Anh, Nguyễn Văn Mậu, Phạm Kim Hùng, Võ Quốc Bá Cẩn, Vasile Cỵrtoaje,… Mình dành nhiều thời gian để biên soạn file nên có sai sót người thơng cảm góp ý cho, trình độ kiến thức có giới hạn Những lời góp ý người gửi vào link email sau mình: thinh06032001@gmail.com; người inbox Facebook Cảm ơn người nhiều! :) P/s: Vì máy bị lỗi nặng nên file PDF bị hết luôn, Mathtype không tương tác máy qua máy tiếng Việt được, ghi tiếng Anh - Đỗ Hữu Đức Thịnh - Season 1: số phương pháp cổ điển đại BĐT kỳ thi Toán Trong season này, trình bày lại BĐT phương pháp tảng thường dùng việc tìm lời giải toán chứng minh BĐT toán cực trị cấp độ khác Ngoài dẫn dắt thêm vài bổ đề với số lời suy luận phát triển thêm (cụ thể ý màu hồng) để người, đặc biệt học sinh đúc kết kinh nghiệm cho giải BĐT I BĐT tương đương (có thể sử dụng biến đổi tương đương để chứng minh): ( a  b) 1) a  b  2ab ( Basic BCS inequality for numbers) 2 a  b2 a  b 2ab   ab   ( Basic RMS  AM  GM  HM inequality for a , b  0) 2 a b  a b  a b a2  b2 2ab   a  b  with a, b 0 such that a  b  2 a b (a  b  c )2 2 2) a  b  c  ab  bc  ca  3(a 2b  b c  c a ) (ab  bc  ca )2 3abc (a  b  c)  bc ca ab     3(a  b  c ) a  b  c  3( ab  bc  ca ) for a, b, c  a b c 3) a  b  c 2ab  2ac  2bc 4) a  b3  c 3abc a, b, c such that a  b  c 0 ( special case : a, b, c 0  Cauchy ' s inequality for non  negative numbers ) 5) (a  b )(c  d ) (ac  bd ) ( Basic Bunyakovsky ' s inequality ) 6) a  b3 ab(a  b) a, b such that a  b 0  4(a  b3 ) (a  b )3 4ab(a  b) for a  b 0 2 4 2 7) 2(a  b  ab) a  b 2ab( 2a  3ab  2b ) a, b  a2  b2 2ab a  ab  b  a  b  with a  b 0 a b a b 2 1  2 with real numbers a, b such that ab  ( special case : a, b  0) a b (a  b ) a b c 9)    with a, b, c  ( Nesbitt ' s inequality )  This is even true for real numbers a; b; c b c c a a b such that ab  bc  ca  8) 10) ( a  b)(b  c)(c  a )  ( a  b  c)(ab  bc  ca ) 8abc for a, b, c 0  abc (b  c  a )(c  a  b )(a  b  c ) for a , b, c are sides of a triangle   (a  b  c)3 27 abc for a, b, c 0 (Cauchy ' s inequality for non  negative variables ) 11) a b (a  b)2   with x, y  ( Basic Bunyakovsky Cauchy  Schwarz inequality ) x y xy a b2  Also, from this inequality we have the chain :   2(a  b ) a  b 2 ab for a, b  b a  2(ax  by ) ( a  b)( x  y ) 2(ay  bx) with a b, x  y 12) ( Basic Chebyshev ' s inequality )  3(ax  by  cz ) (a  b  c)( x  y  z ) 3(az  by  cx ) with a b c, x  y z 13) a 2b  b c  c a ab  bc  ca with a b c  More general : a nb  b n c  c n a ab n  bc n  ca n for a b c 0, n  Z  14) 2(a n  b n ) ( a x  b x )(a y  b y ) with a, b  0; x, y, n  Z  : x  y n  a m n  b mn a m b n  a n b m for a, b 0, m, n  Z  a  ab  b 2 2 15 ) a  ab  b  (a  b) ; a  ab  b  (a  b) a, b    with real numbers a, b : a  b  4 a  ab  b 2 16) a  b  c  d  (a  c)  (b  d ) ( Basic Minkovsky ' s inequality ) 17*) 1   with a, b 0 (The equality happens iff a b 1) 2 (a  1) (b 1) ab  1 18) Consider f (a; b)    with a, b  If ab 1 then f (a; b ) 0; if ab 1 then f (a; b ) 0 2  a  b  ab 1 19)   with a, b  a  b  ab  20) (a  b  c )(a  b  c) 3.max  a 2b  b 2c  c a; ab  bc  ca  with a , b, c 0 21) (a  b  c ) (a  b  c)(a 2b  b c  c a ) a, b, c  a b c b c a       with a b c  a b b c c a a b b c c a 23*) (a  b  c ) 3(a 3b  b3c  c 3a ) (Vasile ' s inequality ) 22) x 1 1  with x 0     a, b, c 0 : a  b  c 3 2 2 1 x 1 a 1 b 1 c 3 3 25) (a  b  c) a  b  c  24abc a, b, c  24) II số BĐT, bổ đề, phương pháp tiếng quen thuộc: (tăng dần theo cấp độ học) a) Thuộc loại THCS – thi lớp 10: a12  a22   an2 a1  a2   an n n 1)   a1a2 an  with  i 1; 2; ; n and n  Z  , n 2 1 n n    a1 a2 an ( RMS  AM  GM  HM inequality for n positive numbers)  If x1 ; x2 ; ; xn are positive real numbers that x1  x2   xn 1 then with same condition for we have : a1 x1  a2 x2   an xn a1x1 a2x2 anxn (We ighted AM  GM inequality) 2) For real numbers ; bi (i 1; 2; ; n) we have :  a12  a22   an2   b12  b22   bn2   a1b1  a2b2   an bn  2 an2  a1  a2   an  a12 a22 (Cauchy  Schwarz inequality )      bi  b1 b2 bn b1  b2   bn xn2 xn2 x12 x22       n  x12  x22   xn2   x1  x2   xn xi  x2 x3 xn x1 3) Let a1 a2  an  If b1 b2  bn then : n(a1b1  a2b2   anbn ) ( a1  a2   an )(b1  b2   bn )  If b1 b2  bn then : n(a1b1  a2b2   anbn ) (a1  a2   an )(b1  b2   bn ) (Chebyshev ' s inequality ) 4) If xi j 0 (i 1; 2; ; n, j 1; 2; ; m) then we have :  ( x1.1  x1.2   x1.n )( x2.1  x2.2   x2.n ) ( xm.1  xm.2   xm.n )  m x1.1 x2.1 xm.1  m x1.2 x2.2 xm.2   m x1.n x2.n xm.n  m ( Holder ' s inequality )  Ex :  (a  b)(c  d )  ac  bd  a, b, c, d 0   (a  b  c)(m  n  p)( x  y  z )  amx  bny  cpz  a, b, c, m, n, p, x, y, z 0 anm a1 a2 a a1m a2m  m  m   m     n for  (i 1; 2; ; n); m  Z  , m 2 a2 a3 a1 a2 a3 a1 a1n  a2n   ann a1  a2   an   with same condition ( Power Mean Inequality ) n n 1 n3 1 n n 1      for  (i 1; 2; ; n)  n  n   n  a1 a2 an (a1  a2   an ) a1 a2 an (a1  a2   an ) n n   Let and bi  (i 1; 2; ; n) and real p, q  such that p  q  pq Then we have : a p p  a   a p n p   b q q q n  b   b  q a1b1  a2b2   anbn  General Holder ' s inequality  5) Let , bi  (i 1; 2; ; n) and any k  Then we have : ( a1k  b1k )(a2k  b2k ) (ank  bnk )  k (a1  a2   an ) k  (b1  b2   bn ) k ( Minkovsky ' s inequality )  similarly for variables , bi , ci  6) For any x  we have :  (1  x) r 1  xr for r 1 and r 0  (1  x) r 1  xr for r 1 ( Bernoulli ' s inequality ) 7) For any positive integer m and a, b, c 0 we have : a m ( a  b)( a  c)  b m (b  c)(b  a)  c m (c  a)(c  b) 0 ( Schur ' s inequality )  This is also true for real m 1 and equality happens iff a b c or (a; b; c) ~ (0; k ; k ) with k   Case m 1  Schur deg : All forms : ( p , q , r will be di scussed in later part )  a  b3  c  3abc ab(a  b)  bc (b  c )  ca (c  a )  (a  b  c )3  9abc 4(a  b  c )(ab  bc  ca ) 9abc 2(ab  bc  ca )  (a  b  c )(a  b  c  ab  bc  ca) 3  ab( a  b)  bc(b  c )  ca (c  a )  a b c  abc (b  c  a )(c  a  b)(a  b  c) (Well  known result )  a2  b2  c   (b  c ) (b  c  a)  (c  a ) (c  a  b)  (a  b) (a  b  c) 0   a  b3  c3   a  b  c    a  b  c   ab  bc  ca   4(a  b3  c )  15abc (a  b  c)3 a b c 4abc     2 b  c c  a a  b (a  b )(b  c )(c  a ) a b c 1     a  b  c    for a, b, c  : abc 1 b c a a b c   a 2b  b c  c a   ab  bc  ca   a 2b  b 2c  c a   ab  bc  ca   Case m 2  Schur deg : All forms :  a  b  c  abc (a  b  c ) ab(a  b )  bc (b  c )  ca (c  a )  a  b  c  2abc (a  b  c) (a  b  c )(ab  bc  ca )  2(ab  bc  ca )  (a  b  c )  6abc(a  b  c) 9abc  a  b  c  ab  bc  ca a  b  c 2  (b  c)(b  c  a )    (c  a)(c  a  b)    (a  b)(a  b  c)  0  Let a, b, c, x, y , z 0 Then we have : x (a  b)(a  c )  y (b  c )(b  a)  z(c  a )(c  b) 0 iff a b c and :  x y  z y  ax  cz by  x  z y  x z y  ax by  cz by (General Vornicu  Schur inequality ) 8) 1    for a, b, c 0, no of which are ( Iran 96 inequality ) (a  b) (b  c ) (c  a ) 4(ab  bc  ca) 9) (a  b  c ) 3(a 3b  b 3c  c3 a) (Vasile ' s inequality )  The equality happens iff a b c and also for 4 2   ( a; b; c)  k sin ; k sin ; k sin  or any cyclic permutation 7 7  10) Let and bi (i 1; 2; ; n) such that :  a1 a2  an 0, b1 b2  bn 0  a1 b1 ; a1  a2 b1  b2 ; ; a1  a2   an  b1  b2   bn   a1  a2   an   an b1  b2   bn   bn For xi 0 we have :  xta11 xta22 xtann  xtb11 xtb22 xtbnn , where (t1; t ; ; t n ) are all the permutations of (1; 2; ; n) sym sym ( Muirhead ' s inequality ) E.g : a  b ab(a  b)  a 3b  b3a a 2b1  a1b a  b  c ab  bc  ca  a 2b  b c  c a a1b1  b1c1  c1a1 a  b  c abc (a  b  c )  a 4b c  b c a  c a 0b a 2b1c1  b 2c1a1  c a1b1 b) Thuộc loại THPT trở lên: (thật khó để viết dạng phát biểu BĐT nên cố vậy.)  a1 a2  an ; b1 b2  bn 1)  If  and (k1; k ; ; kn ) is an arbitrary permutation of (1; 2; ; n)  a1 a2  an ; b1 b2  bn then : a1b1  a2b2   anbn a1bk1  a2bk2   anbkn  If a1 a2  an and b1 b2  bn then : a1b1  a2b2   anbn a1bk1  a2bk2   anbkn ; n  a1b1  a2b2   anbn   a1  a2   an   b1  b2   bn  ( Rearrangement inequality ) 2)  Convex function : If a , b 0 such that a  b 1 then f ( x ) is called a convex function on I (a; b )  R iff x1 ; x2  I we have : f  ax1  bx2  a f  x1   b f  x2  * Concave function : If a, b 0 such that a  b 1 then f ( x ) is called a concave function on I (a; b)  R iff x1; x2  I we have : f  ax1  bx2  a f  x1   b f  x2   If f ( x) is a convex function on interval I  R then for any xi  I (i 1; 2; ; n) we have :  x  x   xn  f  x1   f  x2    f  xn  f (Classic Jensen ' s inequality )  n n    If f ( x) is a convex function on interval I  R then for any xi  I (i 1; 2; ; n) and pi  we have : p1 f  x1   p2 f  x2    pn f  xn   p x  p2 x2   pn xn  f  1  p1  p2   pn p1  p2   pn   And if f ( x) is a concave function then the inequality is reversed (General Jensen ' s inequality )   If f ( x) is a convex and continuous function on interval I  R then for any xi  I (i 1; 2; ; n ) and pi  (0;1) such that p1  p2   pn 1 we have : p1 f  x1   p2 f  x2    pn f  xn   f  p1x1  p2 x2   pn xn  And if f ( x) is a concave function then the inequality is reversed *The classic inequality is a special case from the general one with p1  p2   pn  Let and bi (i 1; 2; ; n)  I ( I  R ) such that : _ a1 a2  an ; b1 b2  bn _ a1 b1 ; a1  a2 b1  b2 ; ; a1  a2   an  b1  b2   bn  _ a1  a2   an  an b1  b2   bn   bn If f ( x) is a convex function on I then we have : f  a1   f  a2    f  an   f  b1   f  b2    f  bn  ( Karamata ' s inequality )  If f ( x) is a convex function on I  R then for  I (i 1; 2; ; n ) we have :  a  a   an  f  a1   f  a2    f  an   n(n  2) f   (n  1)  f  b1   f  b2    f  bn   n   a a  a   an where bi  i  (i 1; 2; ; n) ( Popoviciu ' s inequality ) n n 3) Define p a  b  c, q ab  bc  ca, r abc with a, b, c are any real numbers If k  p  3q then we have : p  pk  2k p  pk  2k r  27 27  The minimum and maximum happens iff of variables a; b; c are equal 4)  Let f (a; b; c) be a symmetric polynomial of degree with a, b, c 0 Then : f (a; b; c) 0  f (1;1;1); f (1;1;0); f (1;0;0) 0 ( SD3 theorem)  Let f (a; b; c) be a cyclic homogeneous polynomial of degree with a, b, c 0 Then : f (a; b; c ) 0  f (1;1;1) 0; f (a; b;0) 0 (CD3 theorem)  Let f n  a; b; c  be a cyclic homogeneous polynomial of degree n (n 3; 4;5) with a , b, c 0 Then f n  a; b; c  0  f n  a;1;1 0 and f n  0; b; c  0 5) ( S O.S technique) Define S S a (b  c )  Sb (c  a )  S c (a  b) , where S a ; S b ; S c are functions with variables a; b; c Then S 0 iff :  Sa ; Sb ; Sc 0  a b c; Sb 0; Sb  S a 0; Sb  Sc 0  a b c; S a 0; S c 0; S a  Sb 0; S c  2Sb 0  a b c; Sb 0; Sc 0; a Sb  b S a 0  Sa  Sb  Sc 0; Sa Sb  Sb Sc  Sc S a 0  Consider f (a; b; c) P (a  b)  Q(a  c)(b  c ) 0 (*)  If f (a; b; c) is symmetric then to prove (*) is true, we assume that a b c or c min  a; b; c or c max  a; b; c and prove that P, Q 0  If f (a; b; c) is cyclic then to prove (*) is true, we assume that c min  a; b; c or c max  a; b; c and prove that P, Q 0 ( S S technique) c) số đẳng thức nâng cao: (*) số đẳng thức thường áp dụng cho toán sử dụng pp S.O.S, S.S: a b (a  b) 1) a  b  2ab (a  b) ;    b a ab 2  n  a  a   a 2 2 n   a a    an   2 n  a12  a22   an2    a1  a2   an  (a  b) 2 For n 2 : 2(a  b )  (a  b)  1i  j n   a j  2(a  b )  a  b For n 3 : 3(a  b  c )  ( a  b  c)  (a  b)  (b  c)  (c  a) 2 2 3(a  b  c )  (a  b  c) 2 (a  b)  (a  c)(b  c) 3(a  b  c )  (a  b  c) (a  b) (b  c) (c  a) 2(a  b)  1   1 1  (a  b  c)              (a  c)(b  c) ab bc ca ab a b c  ac bc  2) ( a  b  c )  3(ab  bc  ca ) a  b  c  (ab  bc  ca )   (a  b)  (b  c)  (c  a )2  (a  b)  (a  c)(b  c) 3) a  b3  ab(a  b) (a  b)(a  b) a b c (b  c) (c  a ) ( a  b) (a  b) (a  b  2c)(a  c )(b  c )          b  c c  a a  b 2(a  b)(a  c) 2(b  c)(b  a ) 2(c  a )(c  b) (c  a )(c  b) 2(a  b)(b  c )(c  a )  3(a  b3  c )  (a  b  c)( a  b  c ) (a  b)(a  b)  (b  c )(b  c )  (c  a )(c  a ) 2( a  b)(a  b)  (a  b  2c)( a  c)(b  c) 4) ( a  b  c)(ab  bc  ca )  9abc (a  b)(b  c)(c  a )  8abc a (b  c )  b(c  a )  c (a  b ) 2c(a  b)  (a  b)(a  c )(b  c) 5) a  b3  c  3abc  (a  b  c)  (a  b)  (b  c)  (c  a)  (a  b  c)  (a  b)  (a  c)(b  c)  a b c (a  b) (a  c)(b  c) 2  (a  b) (3c  a  b)  (b  c) (3a  b  c)  (c  a ) (3b  c  a )   6)      b c a 6abc  ab bc a2 b2 c2 (a  b) (b  c) (c  a )  1  bc 7)    ( a  b  c)       (a  b)  (a  c)(b  c) b c a b c a ac  a b 8) a  b3  c  3abc  ab(a  b)  bc(b  c)  ca(c  a )   (b  c  a )(b  c )  (c  a  b)(c  a )  (a  b  c )(a  b)  (a  b  c)( a  b)  c( a  c)(b  c) a b3 c 9)    ( a  b  c)  (b  c)  (b  c)  a   (c  a )  (c  a )  b   (a  b )2  (a  b)  c  bc ca ab 2abc ( a  b)  c  (a  c)(b  c)   ( a  b)     (a  c)(b  c) abc abc c   a b3 c  a 1 b 1  c 1 10)    (a  b  c )    ( a  b)     (b  c )     (c  a ) b c a  b 2  c 2  a 2 11) a  b  c  a 2b  b 2c  c a  a (b3  c )  b (c  a )  c(a3  b3 )   (a  b )(a  b)  (b  c )(b  c)  (c  a )(c  a)  (a  b )(a  b)  ( a  ab  b  c )(a  c)(b  c) (**) số đẳng thức khác: đẳng thức đầu dùng, đẳng thức từ số 10 trở thường dùng nha a  bc b  ca b  ca c  ab c  ab a  bc   a  b  c  with a b c b c c a c a a b a b b c a  bc b  ca b  ca c  ab c  ab a  bc     a  b  c  with a b c b c c a c a a b a b b c a  b b  c c  a (a  b )(b  c )(a  c ) 2)     a  b b  c c  a (a  b )(b  c )(a  c ) 1)  a  b b  c c  a a (b  c)  b(c  a)  c(a  b)2    with a b c a b b c c a (a  b)(b  c )(a  c ) a b b c b c c a c a a b 3)     with a b c a b b c b c c a c a a b 2 a  b b  c b  c c  a c  a a  b   a (b  c)  b(c  a)  c( a  b)      a b b c b c c  a c  a a b (a  b )(b  c )(c  a )  a  bc b  ca b  ca c  ab c  ab a  bc 4)    a  b  c bc ca c a a b a b b c 2 2 a  bc b  ca b  ca c  ab c  ab a  bc     ( a  b  c) with a b c b c c a c a a b a b b c 5) (a  bc)(b  c )  (b  ca )(c  a )  (c  ab)(a  b) 0 (a  bc)(b  c)  (b  ca )(c  a )  (c  ab)( a  b)  2( a  b)(b  c)(c  a ) (b  c) (c  a ) ( a  b) 6)   1 with a b c (a  b)(a  c) (b  c)(b  a ) (c  a )(c  b)  ab  bc  bc  ca  ca  ab 7)    with a b c a b b c b c c a c a a b a  bc b  ca b  ca c  ab c  ab a  bc 8)    a  b  c with a b c b c c a c a a b a b b c 2 2 a  bc b  ca b  ca c  ab c  ab a  bc     ( a  b  c)  ab( a  b)  bc(b  c)  ca(c  a)  bc ca c  a a b a b b c a  bc b  ca b  ca c  ab c  ab a  bc 9)    a  bc b  ca b  ca c  ab c  ab a  bc 10) (a  b  c )3 a  b  c  3(a  b)(b  c )(c  a ) 11) 2(a 2b  b c  c a )  (a  b  c ) (a  b  c)(b  c  a )(c  a  b)(a  b  c ) 12)  ab n  bc n  ca n    a nb  b n c  c n a  (a  b )(b  c )(c  a ) a pb q c r with p , q , r  N and n 2 s uch that p  q  r n  (?) sym E.g : for n 2 :  ab  bc  ca    a 2b  b 2c  c a  (a  b)(b  c )(c  a ) n 3 :  ab3  bc  ca    a 3b  b 3c  c 3a  (a  b)(b  c )(c  a )(a  b  c ) n 4 :  ab3  bc3  ca    a 3b  b3c  c a  ( a  b)(b  c)(c  a)(a  b  c  ab  bc  ca ) 13) a n (a  b)(a  c )  b n (b  c )(b  a )  c n (c  a )(c  b)    a n  bn  c n  ( a  b)   b n  c n  a n  (b  c )   c n  a n  b n  (c  a )  1 14) If a, b, c 0 such that abc 1 then :   1 ab  b  bc  c 1 ca  a 1 2a  bc 2b  ca 2c  ab 15) a  b  c  abc 4  2a  bc  (4  a )(4  b ), etc    (2  b)(2  c) (2  c)(2  a) (2  a)(2  b) (2  b)(2  c) (2  c)(2  a ) (2  a )(2  b) a b c bc ca ab          1 2a  bc 2b  ca 2c  ab 2a  bc 2b  ca 2c  ab 2a  bc 2b  ca 2c  ab 1 1 a b c 2       (a  b  c  2) (2  a )(2  b)(2  c) 2a  bc 2b  ca 2c  ab a  b  c  2(ab  bc  ca)  abc  From the identity , there exists x, y , z  such that : a 2 xy yz zx ; b 2 ; c 2 ( z  x)( z  y ) ( x  y )( x  z ) ( y  z )( y  x) And there exists triangle ABC such that : a 2 cos A; b 2 cos B; c 2 cos C a  bc b  ca c  ab 16) a  b  c  2abc 1  a  bc  (1  a )(1  b ), etc     (1  b)(1  c) (1  c)(1  a) (1  a)(1  b) (1  b)(1  c) (1  c)(1  a ) (1  a)(1  b) a b c   1    2 a  bc b  ca c  ab a  bc b  ca c  ab bc ca ab 1 a  b  c 1    1       ( a  b  c  1) 2(1  a)(1  b)(1  c ) a  bc b  ca c  ab a  bc b  ca c  ab a  b  c  ab  bc  ca  abc a b c  If we substitute a  ; b  ; c  we will get identity 15, so : 2 xy yz zx From the identity , there exists x , y , z  such that : a  ;b ;c ( z  x)( z  y ) ( x  y )( x  z ) ( y  z )( y  x) And there exists triangle ABC such that : a cos A; b cos B; c cos C  Also if we let a  yz ; b zx; c  xy with x, y , z  then abc  xyz , so we have identities 17  18 : 17) xy  yz  zx  xyz 4  y xyz 1 x y z x z      1     x2 y2 z 2 x2 y2 z 2 2 x 2 y 2 z xy  yz  zx  18) xy  yz  zx  xyz 1  y xyz 1 x y z x z   2    1     x  y 1 z 1 x 1 y 1 z 1 1 x 1 y  z xy  yz  zx   From identity 17, there also exists m, n, p  such that x  2m 2n 2p ; y ;z , similarly for identity 18 n p pm mn 19) x  x  xy  y   y  y  yz  z   z  z  zx  x   y  x  xy  y   z  y  yz  z   x  z  zx  x  ( x  y  z )( x  y  z z  x  xy  y   x  y  yz  z   y  z  zx  x  ( x  y  z )( xy  yz  zx ) 20)  x  xy  y   y  yz  z   z  zx  x   p  pq  q , where p x y  y z  z x; q xy  yz  zx 3  x y  y z  z x   xy  yz  zx    ( x  y )( y  z )( z  x)  2  xy ( x  y)  yz ( y  z )  zx( z  x)    ( x  y )( y  z )( z  x) 4   ( x  y  z )  x y  y z  z x   ( xy  yz  zx)  x  y  z      ( xy  yz  zx)  x  y  z  ( x  yz )( y  zx )( z  xy )   ( x  y )( y  z )( z  x)   ( x  y  z )  x y  y z  z x  (2 x  yz )(2 y  zx )(2 z  xy )   ( x  y )( y  z )( z  x)  2 21) 2( x  y )( y  z )( z  x )  xy ( x  y )  yz ( y  z )  zx( z  x)  xyz    ( x  y )( y  z )( z  x)  2 22)  x ( y  z )  y ( z  x)  z ( x  y )   x ( y  z )  y( z  x)  z( x  y)    ( x  y )( y  z )( z  x)  a2 b2 c2  ab  bc  ca  2 23)      abc  a bc or b ca or c ab a  bc b  ca c  ab  a b c  d) số bổ đề BĐT phụ thường dùng: Ở phần giới thiệu số bổ đề rồi, phần bổ sung thêm vài bổ đề phụ theo kinh nghiệm học tìm hiểu từ thầy tồn giới, nhiều đáng mà  Nếu cảm thấy thiếu bổ sung thoải mái, theo kinh nghiệm thơi  BĐT với điều kiện biểu thức p,q,r – phần 1: Trong phần cho tốn có điều kiện quen thuộc, phần điều kiện “khơng bình thường” trình bày phần sau d 1) If a, b, c  such that abc 1 then : 1)a m  b m  c m a n  b n  c n (m, n  Z  ; m  n) a b c a b c 1 a b c a b c 1 2)   a  b  c;          (a  b  c  1)     a  b  c    b c a b c a a b c b c a b c a a b c 1 3) k  2k  2k 1 with k  Z  k k k a  a  b  b 1 c  c 1 1 4) k   1 with k  Z  k k k k k a  b  b  c  c  a 1 1 5)     a(b  1) b(c  1) c(a  1) 1 1         with m, n  (*) a (a  b) b(b  c ) c (c  a ) a (ma  nb) b(mb  nc ) c( mc  na ) m  n 1 1 6)    a  2b  b  2c  c  2a  ab bc ca 7) 5  5  1 a  b  ab b  c  bc c  a  ca 8) Let p a  b  c; q ab  bc  ca then : 4 q q p  4 q p p  p  4q   pq 5 p  a b c a b c a b c 1   1       1    b  c  c  a 1 a  b  b  c  c  a 1 a  b 1 a  b  c  a2 b2 c2 1 1 1  1 1     1    Extra :         2a  2b  2c  a 2 b2 c 2  3a  3b  3c  a  b  c   9) 10) (a  1)(b  1)(c  1) ( a  b)(b  c )(c  a) (a  1)(b  1)(c  1) 8 11) ( n  1) a   (n  1)b   (n  1)c   n(a  b  c ) with a  b  c ab  bc  ca 1  Special case : a  b  c    and without condition abc 1 a b c 12) 1    (a  1) (b  1) (c  1) 13) (a  b  c )5 81 a  b  c  d 2) If a, b, c  such that a  b  c 3 then : 1) a  b  c 3 ab  bc  ca 2) a m  b m  c m 3 for m  Z  am bm cm 3) n  n  n 3 for m, n  Z  such that m n b c a 1 4)   a  b  c a b c a n  bn bn  cn c n  a n n n n 5) a  b  c    a n  b n  c n with all n  Z  a b b c ca 1 6)    with all n 2 a  b2  n b2  c2  n c  a  n n  am bm cm 7) n n  n   with m, n  Z  such that m  n n n n b c c a a b 1 1 1 n 8) abc  a  b  c  3   abc   a  b  c  3 with n  Z        a  b  c a b c ab bc ca 1 9)   1 a  ab  b b  bc  c c  ca  a 10) a 2b  b c  c a  abc 4 11) a  b  c a 2b  b 2c  c a a b c 12*)    with p 1 b  pc c  pa a  pb  p d 3) If a, b, c  : ab  bc  ca 3 then : 1) a  b  c 3abc  (a  b  c )5 243abc a b c 2)   a  b  c b c a 1 1 3)        a  b 1 c 1 (a  1) (b  1) (c  1) 4) a 2b  b c  c a a  b  c 1 5)    with m 2 2 (a  b)  m (b  c)  m (c  a )  m  m 1     with x, y  and m 2  x  xy  y  (?) 2 2 ( xa  yb)  m ( xb  yc)  m ( xc  ya)  m ( x  y )  m 1 6)    with k 1 a  b2  k b2  c  k c2  a  k  k 1 7)   1 a 2 b 2 c 2 1 8*)   1 a  b  b  c 1 c  a 1 d 4) If a, b, c  such that a  b  c 3 then : 1) a  b  c ab  bc  ca  n a  n b  n c ab  bc  ca with n  Z  a b c 2)     (a  b  c)3 9(ab  bc  ca) b c a a b c a b c a b c 3)   1;    b2 c2 a2 b  c 1 a 1 4) a 3b  b3c  c 3a 3  a b c 1 5) a 2b  b c  c a 2  abc  In case a, b, c 0 : the equality happens iff   a  2; b 1; c 0 or any cyclic permutation d 5) If a, b, c  such that a  b  c ab  bc  ca then : 1   1 1 a  b 1 b  c 1 c  a  a  b  c  4abc 2) a  b  c  abc 4    (a  1)(b  1)(c  1) 8 1) 1    a b b c c a 4) (a  b)(b  c )(c  a ) 8 3) a b2 c a b2 c2 3n 2 5)   a  b  c     3  n with n 3 b c a b c a a  b2  c2 6) a  b  c abc  d 6) If a, b, c  such that ab  bc  ca  abc 4 then : a b c a b c 1) a  b  c ab  bc  ca 3     b c c a a b a b c    a  b  c b c a a b c 1 a b c 1 2)         a  b 1 c  2 a b b c c a 1 3)   1 a 2 b 2 c 2 4) ab  bc  ca 3 1 5)   a  b  c a b c 6) a 2b  b c  c a a  b  c d 7) If a, b, c  such that a  b  c  abc 4 then : 1) ab  bc  ca a  b  c 3  ab  bc  ca abc  a  b  c 3 1 2)     a  b b  c c  a 2(ab  bc  ca ) 3) a (b  c )  b(c  a )  c(a  b ) 6 bc ca ab bc ca ab 4)   3    a  b  c a b c a b c  bc  5) a  bc a    2; etc   6) a  b  c  ab  bc  ca 2(a  b  c)  a  b  c  a  b  c ab  bc  ca a b c d 8) If a, b, c  such that a  b  c 2(ab  bc  ca) then :  2abc  the equality happens iff (a; b; c ) ~ (k ; k ; 4k ) with k  d 9) If a, b, c  such that ab  bc  ca abc  then : 1) Assume that (b  1)(c  1) 0 then : c  ab 2  a  b  c  abc 4 2) Max  ab; bc; ca 1; Max  a; b; c 1  BĐT với điều kiện “cổ điển – ngắn gọn” (như biến dương không âm,…): a b c a b c    b c a abc 1 n d 11) If a1; a2 ; ; an  then :      a1n  a2n  ann  a1a2 an d 10) If a, b, c  then : a n  bn bn  c n cn  a n a n  bn  cn d 12) If a, b, c 0, no of which are then :   3 with n  Z  a b b c ca a b c a  b2  c2 a b c a2  b2  c2 d 13) If a, b, c  then :      1 ab  bc  ca b  c c  a a  b 2(ab  bc  ca ) a b c a b c d 14) If a, b, c  then :     (b  c )2 (c  a ) (a  b) ab  bc  ca 4(a  b  c) d 15) If a, b, c, x, y , z  then : a ( y  z )  b( z  x)  c ( x  y ) 2 (ab  bc  ca )( xy  yz  zx) d 16) If a, b, c, d 0 then : a  b  c  d  2abcd a 2b  a c  a d  b c  b d  c d (Turkevich ' s inequality )  The equality happens iff a b c d and a b c k  0, d 0 or any cyclic permutation d 17) If a, b, c  then : a b b c c a b c   a   4     c a b  b c c a a b  d 18) If a, b, c are sides of a triangle th en : 3(a  b  c )  a  b  c  b  c  a  c  a  b  a  b  c d 19) If a, b, c  then : a3 b3 c3 a2  b2  c a  b  c      b  bc  c c  ca  a a  ab  b a b c a3 b3 c3 3(ab  bc  ca )     a  b  c  b  bc  c c  ca  a a  ab  b a b c a b c a b c a b c d 20) If a, b, c  then :       b  bc  c c  ca  a a  ab  b ab  bc  ca a  2bc b  2ca c  2ab a b c a  b2  c2 d 21) If a, b, c  then :   3 b c a ab  bc  ca x y z d 22) If a, b, x, y, z  then :    ay  bz az  bx ax  by a  b n x2 x  d 23*) If x 0 then : e 1  x  1  x  e x     n n ai3 a  a   an d 24) If  i 1; 2; ; n then :    an1 a1  i 1  1  1 x d 25) If a, b, c  then : a b c 2ab    2 b  c c  a a  b ( a  b) d 26) If a, b, c  then : a  b  c  2abc  2(ab  bc  ca )  a  b  c  abc  3(a  b  c) d 27*) If a, b, c are sides of a triangle then : a 2b(a  b)  b 2c(b  c)  c a (c  a) 0 ( IMO 1983) am bm cm a m bm c m a b3 c a b c d 28) If a, b, c  and m  Z then : m   m  m   m  m  m         a  b  c b c a b c a b c a b c a a b c d 29) If a, b, c  then :   1 a  (a  b)(a  c) b  (b  c )(b  a ) c  (c  a )(c  b )  d 30) If a, b, c  then : a b c    a b bc ca d 31) If a, b, c are sides of a triangle then : a  b  c 2  a 2b  b c  c a  (the equality happens iff a, b, c are sides of a degenerate triangle.) a b c ka  b kb  c kc  a d 32*) If a, b, c  and k 0 then :      b c a ka  c kb  a kc  b a b c 8abc d 33) If a, b, c  then :    4 b c a (a  b)(b  c)(c  a ) a  b2  c2 8abc d 34) If a, b, c 0, no of which are then :  2 ( Jack Garfunkel ' s inequality ) ab  bc  ca (a  b)(b  c)(c  a )  BĐT với điều kiện “cổ điển – ngắn gọn” – phần 2: Trong phần trình bày số BĐT đại lượng p,q,r (với quy ước p = a+b+c; q = ab+bc+ca; r = abc), kể BĐT Schur Trước tiên ta có số đẳng thức liên hệ p,q,r sau: 1) a  b  c  p  2q 2) a 2b  b 2c  c a q  pr 3) (a  b)(b  c)(c  a )  pq  r  ab(a  b)  bc(b  c )  ca (c  a)  pq  3r 4) (a  b)(a  c)  (b  c)(b  a )  (c  a)(c  b)  p  q 5) a  b3  c  p  pq  3r 6) a 3b3  b3c  c 3a q  pqr  3r 7) a  b  c  p  p q  2q  pr 8) ab(a  b )  bc(b  c )  ca (c  a )  p q  2q  pr 9) Denote S  p 3r  p q  18 pqr  4q  27 r , then :   (a  b)(b  c)(c  a)  S pq  3r  S pq  3r  S if (a  b)(b  c )(c  a)  0; if ( a  b)(b  c)(c  a ) 0 2 p q  2q  pr  p S p q  2q  pr  p S 3  a b b c c a  if (a  b)(b  c)(c  a )  0; if (a  b)(b  c )(c  a ) 0 2 10) a 2b (a  b )  b c (b  c )  c a (c  a )  p 3r  p q  pqr  2q  3r  a 2b  b c  c a  11) ab(a  b )  bc (b  c )  ca (c  a )  p q  p 3r  p q  pqr  2q  3r 12) a  b6  c  p  p q  p 3r  p q  12 pqr  2q  3r  số BĐT mối liên hệ đại lượng p,q,r: 1) p 3q; q 3 pr 2) pq 9r 27 9q  p  ( pq  r ) 3 pq  27 r p  p(4q  p )  p (4q  p ) 3) p  9r 4 pq  r   r max 0;  9   4)2 p  9r 7 pq 5) p q  pr 4q 6) p  4q  pr 5 p q  (4q  p )( p  q )  (4q  p )( p  q) 7) r   r max 0;  6p 6p   p (5q  p ) p  p q  13q 8) r  ; r 18 9p  p (5q  p ) p  p q  13q   p (4q  p ) (4q  p )( p  q )  Combining inequality 3, 6, and we get :  ; ;  r max 0;  18 9p 6p     pq  p  k k pq  p  2k k 9*) r  , with k  p  3q This result comes from solving the inequality 27 27 S  p r  p q  18 pqr  4q  27r 0 with variable r  From this result , by AM  GM we have : 3  27 r 9 pq  p  2k k 9 pq  p  2( p  3q)( p  2q) p  p  3q  p ( p  2q ) p (9q  p )( p  2q )  ( p  3q)   p  2q   p  p  3q   ( p  2q )   p  14 p q  6q   p ( p  3q)     p ( p  2q ) p ( p  2q )  (4q  p )  p  10 p q  3q  p ( p  2q )   2( p  3q )  p  q  p  p  3q     27 r 9 pq  p  2k k 9 pq  p    p  p2  q      2  2 p  pq  p   p  3q   2( p  3q )   p  q   p  p  3q   2 2 2   2 p  q 2(9 pq  p )  ( p  q )(2 p  q )  p p  q             2 p  p  3q  p  p  3q  2p   3q    p  p q  9q   p  p  p q  9q  p  p  3q  27 q ( p  q)  p  p  3q  (4q  p )  p  10 p q  3q  q2 q ( p  q) So we have the chain :  r  , more interesting , the third inequality is stronger than p p  p  3q  27 p( p  2q) Schur deg and 4, since : (4q  p )  p  10 p q  3q  27 p( p  2q) (4q  p )  p  10 p q  3q  27 p( p  2q) p (4q  p ) (4q  p )( p  3q)( p  q)   0 in case 4q  p 2 27 p ( p  2q) (4q  p )( p  q) (4q  p ) ( p  3q)   0 6p 54 p( p  2q )  (4q  p )  p  10 p q  3q   Hence we can also conclude that : r max 0;  27 p ( p  q )   p  pk  2k k p  pk  2k k Further more, we can write inequality as : r  27 27   p k  27 p k  r  p k   p2 k  27 Theo nghĩ phương pháp p,q,r phương pháp khó dùng thời gian phân tích, xử lý tính tốn lâu đòi hỏi tính cẩn thận cao, phương pháp đẹp hữu ích việc giải toán (chứng minh + cực trị) khó biểu thức đối xứng – hốn vị Có phương pháp có nhiều điểm tương đồng với phương pháp p,q,r phương pháp u,v,w (có thể tra mạng để tìm hiểu)  ...  ca c  ab 16 ) a  b  c  2abc 1  a  bc  (1  a ) (1  b ), etc     (1  b) (1  c) (1  c) (1  a) (1  a) (1  b) (1  b) (1  c) (1  c) (1  a ) (1  a) (1  b) a b c   1    2 a... c a a1b1  b1c1  c1a1 a  b  c abc (a  b  c )  a 4b c  b c a  c a 0b a 2b1c1  b 2c1a1  c a1b1 b) Thuộc loại THPT trở lên: (thật khó để viết dạng phát biểu BĐT nên cố vậy.)  a1 a2... b  c   9) 10 ) (a  1) (b  1) (c  1) ( a  b)(b  c )(c  a) (a  1) (b  1) (c  1) 8 11 ) ( n  1) a   (n  1) b   (n  1) c   n(a  b  c ) with a  b  c ab  bc  ca 1  Special case