Tuyển tập đề thi vô địch bất đẳng thức thế giới P5 , tài liệu tham khảo, tài liệu gồm các bài toán bất đẳng thức cực khó, với bài tập hay và cách giải được sưu tầm trên thế giói các bạn có đào sâu kiến thức toán về mảng này, Tai liệu được viết bằng tiêng anh. Chúc các bạn học tốt. | Old and New Inequalities 111 and so n n n nan 1 ai an 1 N1 n l a i E ai i l i l i l Using this last inequality it remains to prove that E 1 - E -h E - E 1 i l i l 1 1 2 1 n i I di n l an 1 an 1 j 0. j i Now we will break this inequality into n JI ai n l an l an l j 0 i l and n n n n n E 1 - E -r E - E - 2 1 2 1 2 1 2 1 Let us justify these two inequalities. The first one is pretty obvious n n ai n 1 an_ i an_i_1 n i n i n 1 an i an i 2 1 2 1 n a 1 - an i n - l a 1 - a - 0. 2 1 Now let us prove the second inequality. It can be written as n n n n E 1 -EnE -E r1 2 1 2 1 2 1 2 1 n n i Because n E -E - 1 0 using Chebyshev s Inequality and dn t it i i i i n is enough to prove that n n A n n 2 1 2 1 2 1 2 1 but this one follows because na 1 a -1 af for all i. Thus the inductive step z n 1 1 is proved. 117. Prove that for any Xi X2 xn 0 with product 1 n E -xf 2 E -n- l 2 j n 2 1 A generalization of Turkevici s inequality 112 Solutions Solution Of course the inequality can be written in the following form n n 2 52 xk 12 l fc l A 1 We will prove this inequality by induction. The case n 2 is trivial. Suppose the inequality is true for n 1 and let us prove it for n. Let n n fai x2 . zn -fa 1 I 52xfc I 52Xfc I A 1 c l and let G n x2X3. xn where we have already chosen xi min xi x2 . xn . It is easy to see that the inequality xi x2 xn xi G G . G is equivalent to n n n n - 1 2 xk - I 52 Xk 2xi I 52 _ n I k 2 t 2 t 2 n Clearly we have Xi G and 52 xk n 1 G so it suffices to prove that fc 2 n n n n -1 52 xl - i 52Xk - 2g 52Xk k 2 k 2 A 2 We will prove that this inequality holds for all x2 . xn 0. Because the inequality is homogeneous it is enough to prove it when G 1. In this case from the induction step we already have and so it suffices to prove that 52 xk n - 1 52 2xk 52 Xfc -k 2 k 2 k 2 clearly true. Thus we have proved that xi x2 . xn xi G G . G . Now to com plete the inductive step we will prove that xi G G . G 0. Because clearly Xi the last assertion reduces to proving that -
