设a,b∈R+,求证(ab)^(a+b)/2≥a^b b^a

就是求证
lg (ab)^(a+b)/2 >=lg a^b b^a
(a+b)/2 *(lga+lgb) >=blga+algb
(a+b)*(lga+lgb)>=2blga+2algb
alga+blga+algb+blgb>=2blga+2algb
alga+blgb>=blga+algb
lga^a+lgb^b>=lga^b+lgb^a
就是求证 a^a *b^b>=a^b*b^a
a^a/a^b*b^b/b^a>=1
a^(a-b)*b^(b-a)>=1
a^(a-b)/b^(a-b)>=1
(a/b)^(a-b)>=1
(1)a=b 则上式成立
(2)a>b a/b>1 a-b>0 所以(a/b)^(a-b)>=1成立
(3)a