Question

已知a>b>0,求证a5+b5>a4b+ab4 （2）已知a>2,b>2证明ab>a+b

Answer 1

1. a5+b5-(a4b+ab4 )
=(a^5-a^4b)+(b^5-ab^4)
=a^4(a-b)-b^4(a-b)
=(a-b)(a^4-b^4) a>b>0 a^4>b^4>0
>0
所以
a5+b5>a4b+ab4

2. a>2,b>2 ab>4

2ab-2a-2b=ab-2a-2b+ab>ab-2a-2b+4=(a-2)(b-2)>0
所以 2ab-2a-2b>0
ab>a+b

Answer 2

(a^4-b^4)(a-b)>0
a^5+b^5>a^4b+b^4a

(a-1)(b-1)>1
ab>a+b

Answer 3

[[[[[1]]]]]
用"差法"
[(a^5)+(b^5)]-[(a^4)b+a(b^4)]
=(a-b)(a^4)-(a-b)(b^4)
=(a-b)(a^4-b^4)
∵a＞b＞0
∴a-b＞0且a^4-b^4＞0
∴(a-b)(a^4-b^4)＞0
∴原不等式成立.
[[[[[2]]]]
∵a＞2, b＞2
∴a-1＞1, b-1＞1
∴两式相乘,可得
(a-1)(b-1)＞1
展开,整理可得:
ab＞a+b