1 Vietnam Inequality Forum www.batdangthuc.net Articles Written by Member: Pham Kim Hung User Group: Admin This product is created for educational purpose. Please don’t use it for any commecial purp ose unless you got the right of the author. Please contact www.batdangthuc.net for more details. 2 Dien Dan Bat Dang Thuc Viet Nam www.batdangthuc.net Tac Gia Bai Viet: Pham Kim Hung Nhom: Admin Bai Viet Nay (cung voi file PDF di kem) duoc tao ra vi muc dich giao duc. Khong duoc su dung ban EBOOK nay duoi bat ky muc dich thuong mai nao, tru khi duoc su dong y cua tac gia. Moi chi tiet xin lien he: www.batdangthuc.net. 3 Article Written by Pham Kim Hung Stanford University Since this following theorems are useful to solve recent problem, I decide to share them for every Mathlinkers from my book. Detailed feature and more applications will be shown in this book published in Gil 2007 (IMO). Notice that this is not in Vietnamese version of 2006. Theorem 1. Suppose that F is a homogeneous symmetric polynomial of n variables x 1 ,x 2 , , x n such that deg f ≤ 3. Prove that the inequality F (x 1 ,x 2 , , x n ) ≥ 0 holds if and only if F (1, 0, 0, , 0) ≥ 0; F (1, 1, 0, 0, , 0) ≥ 0; F (1, 1, 1, , 1, 0) ≥ 0; F (1, 1, 1, , 1) ≥ 0; Proof. We denote t = x 1 +x 2 2 ,x= x 1 ,y = x 2 and F = a n i=1 x 3 i + b n i<j x i x j (x i + x j )+c i<j<k x i x j x k ; Denote A = n i=3 x j ; B = n i=3 x 2 j ; C = 2<i<j x i x j ; W e have F (x 1 ,x 2 , , x n ) − F (2t, 0,x 3 , , x n ) = a x 3 + y 3 − (x + y) 3 + bxy(x + y)+b x 2 + y 2 − (x + y) 2 A + xyA = xy (−3a(x + y)+b(x + y) − 2bA + A); and, alternatively, we have F (x 1 ,x 2 , , x n ) − F (t, t, x 3 , , x n ) = a x 3 + y 3 − (x + y) 3 4 + b(x + y) xy − (x + y) 2 4 +b x 2 + y 2 − (x + y) 2 2 A + A xy − (x + y) 2 4 = (x − y) 2 4 (3a(x + y)+−b(x + y)+2bA − A); 4 According to two above relationships, we may conclude that at least one the two following inequalities hold F (x 1 ,x 2 , , x n ) ≥ F (2t, 0,x 3 , , x n )orF (x 1 ,x 2 , , x n ) ≥ F (t, t, x 3 , , x n ); Let t = 1 n (x 1 + x 2 + + x n ). According to UMV theorem (AC theorem), we conclude that the inequalityF (x 1 ,x 2 , , x n ) ≥ 0 holds if and only if F (t, 0, 0, , 0) ≥ 0; F (t, t, 0, 0, , 0) ≥ 0; F(t, t, t, , t, 0) ≥ 0; F (t, t, t, , t) ≥ 0; Since the inequality is homogeneous, we have the desired result immediately. ∇ Theorem 2 (”Symmetric inequality of Degree 3” theorem - SID theorem). Consider the following symmetric expression (not necessarily homogeneous) F = a n i=1 x 3 i +b i<j x i x j (x i +x j )+c i<j<k x i x j x k +d n i=1 x 2 i +e i<j x i x j +f n i=1 x i +g. For t = x 1 +x 2 + + x n , the inequality F ≥ 0 holds for all non-negative real numbers x 1 ,x 2 , , x n if and only if F t n , 0, , 0 ≥ 0; F t n , t n , 0, , 0 ≥ 0; F t n , t n , , t n , 0 ; F t n , t n , , t n ≥ 0; Proof. We will fix the sum x 1 + x 2 + + x n = t = const and prove that the inequality F ≥ 0 holds for all t>0. Indeed, we can rewrite the expression F to (with the assumption that x 1 + x 2 + + x n = t) F = a n i=1 x 3 i + b i<j x i x j (x i + x j )+c i<j<k x i x j x k + + n i=1 x i d t n i=1 x 2 i + e t i<j x i x j + f t 2 n i=1 x i 2 n i=1 x i + g t 3 n i=1 x i 3 . Notice that in this expression, t is an constant. Since this representation of F is symmetric and homogeneous, we may conclude that F ≥ 0 if and only if (according to the above proposition) F t n , 0, , 0 ≥ 0; F t n , t n , 0, , 0 ≥ 0; F t n , t n , , t n , 0 ; F t n , t n , , t n ≥ 0; and the desired result follows. ∇ . inequality is homogeneous, we have the desired result immediately. ∇ Theorem 2 (”Symmetric inequality of Degree 3” theorem - SID theorem) . Consider the following symmetric expression (not necessarily homogeneous) F. Moi chi tiet xin lien he: www.batdangthuc.net. 3 Article Written by Pham Kim Hung Stanford University Since this following theorems are useful to solve recent problem, I decide to share them. 1 Vietnam Inequality Forum www.batdangthuc.net Articles Written by Member: Pham Kim Hung User Group: Admin This product is created for educational purpose. Please don’t use