菜鸟数学题（完）

由a>2得
loga(a+1)>0,loga(a-1)>0

由当a=b,a>0,b>0时，a*b<(a^2+b^2)/2 得

loga(a+1)*loga(a-1)
<[loga(a+1)+loga(a-1)]^2/4
=[loga(a^2-1)]^2/4
<[loga(a^2)]^2/4
=2^2/4
=1

a>2,所以loga(a-1)〉0，loga(a+1)〉0
且loga(a-1)不等于loga(a+1)
所以loga(a-1)*loga(a+1)<{[loga(a-1)+loga(a+1)]/2}^2
=[loga(a^2-1)/2]^2
因为a>2,所以loga(a^2-1)/2〉0
0<loga(a^2-1)/2=loga√(a^2-1)<loga√a^2=1
所以[loga(a^2-1)/2]^2<1
即loga(a-1)*loga(a+1)<[loga(a^2-1)/2]^2<1

loga(a-1)+loga(a+1)=loga(a^2-1)<loga(a^2)=2

而2根号[loga(a-1)*loga(a+1)]<loga(a-1)+loga(a+1)<2

两边平方就得到loga(a-1)+loga(a+1)<1