請問這題數學怎麼算?

Question

請問數學先知,以下1題數學該如何計算?

A=( 999*999*999*.....乘以111個 +1 ) / ( 999*999*999*.......乘以222個 +1 )
B=( 999*999*999*.....乘以222個 +1 ) / ( 999*999*999*.....乘以333個 +1 )
答：A>B

Answer 1

令x=999,y=111

所以
A=( 999*999*999*.....乘以111個 +1 ) / ( 999*999*999*.......乘以222個 +1 )
=(x^y +1)/(x^2y +1)
=[(x^y +1)(x^3y +1)]/[(x^2y +1)(x^3y +1)]

B=( 999*999*999*.....乘以222個 +1 ) / ( 999*999*999*.....乘以333個 +1 )
=(x^2y +1)/(x^3y +1)
=[(x^2y +1)(x^2y +1)]/[(x^2y +1)(x^3y +1)]

A的分子部分為
[(x^y +1)(x^3y +1)]=x^4y +(x^y +x^3y) +1

B的分子部分為
[(x^2y +1)(x^2y +1)]=x^4y +2x^2y +1

所以A,B只要比較(x^y +x^3y)和2x^2y 哪一個大就可

又
x^y +x^3y≧2√(x^y ×x^3y)=2x^2y
(科西不等式)
因
x^y ≠x^3y
所以
x^y +x^3y＞2√(x^y ×x^3y)=2x^2y

所以 A>B #

2006-06-19 22:49:03 補充：
言重,言重.