证:a5+b5>=a3b2+a2b3,a,b>0
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证:a5+b5>=a3b2+a2b3,a,b>0
证:a5+b5>=a3b2+a2b3,a,b>0
证:a5+b5>=a3b2+a2b3,a,b>0
a^5-a^3b^2+b^5-a^2b^3
=a^3(a^2-b^2)+b^3(b^2-a^2)
=(a^2-b^2)(a^3-b^3)
=[(a+b)(a-b)[(a-b)(a^2+ab+b^2)]
=(a+b)(a-b)^2[(a+b/2)^2+3b^2/4]
a>0,b>0
所以a+b>0
(a-b)^2>=0
(a+b/2)^2>=0,3b^2/4>=0
所以(a+b/2)^2+3b^2/4>=0
若取等号则a+b/2=0,b=0
和b>0矛盾
所以(a+b/2)^2+3b^2/4>0
所以(a+b)(a-b)^2[(a+b/2)^2+3b^2/4]>=0
仅当a-b=0,a=b时取等号
所以a^5-a^3b^2+b^5-a^2b^3>=0
所以a^5+b^5>=a^3b^2+a^2b^3
5(a+b)>=ab(6+6)
a5+b5-a3b2-a2b3
=a3(a2-b2)+b3(b2-a2)
=(a2-b2)(a3-b3)
=(a+b)(a-b)(a-b)(a2+ab+b2)
=(a+b)(a-b)2(a2+ab+b2)>=0
证毕
a^5+b^5-a^3b^2-a^2b^3
=a^3(a^2-b^2)-b^3(a^2-b^2)
=(a^3-b^3)(a^2-b^2)
=(a-b)(a^2+ab+b^2)(a+b)(a-b)
=(a-b)^2(a+b)(a^2+ab+b^2)
显然大于等于0!
a5+b5-a3b2-a2b3=
a3(a2-b2)-b3(a2-b2)=
(a3-b3)(a2-b2)>=0
无论a>b or a