设a=2,b=1/2
a^2+b^2
=(a+b)^2-2ab
=(a+b)^2-2
=(5/2)^2-2
=(25/4)-2
=17/4>0
2根号2(a-b)
=2根号2(2-1/2)
=2根号2(3/2)
=3根号2>0
17/4=4.25
3*1.414=4.242
17/4>3根号2
所以a^2+b^2>2根号2(a-b)
如果a>b且ab=1,比较a^2+b^2与2*2^2(a-b)的大小
设a=2,b=1/2
a^2+b^2
=(a+b)^2-2ab
=(a+b)^2-2
=(5/2)^2-2
=(25/4)-2
=17/4>0
2根号2(a-b)
=2根号2(2-1/2)
=2根号2(3/2)
=3根号2>0
17/4=4.25
3*1.414=4.242
17/4>3根号2
所以a^2+b^2>2根号2(a-b)