已知a>b>0,求证a^ab^b>(ab)^[(a+b)/2]

已知a>b>0,求证a^ab^b>(ab)^[(a+b)/2]
已知a>b>0,求证a^ab^b>(ab)^[(a+b)/2]

已知a>b>0,求证a^ab^b>(ab)^[(a+b)/2]
a^ab^b/{(ab)^[(a+b)/2]}
=a^[(a-b)/2]*b^[(b-a)/2]
=(a/b)^[(a-b)/2]
∵ a>b>0
∴ a/b>1 a-b>0
则 (a/b)^[(a-b)/2]>=(a/b)^0=1
故 (a/b)^[(a-b)/2]>1
即 a^ab^b>(ab)^[(a+b)/2]