证:令a/b=x ≥ 1,变形得:lnx ≥ 2(x-1)/(x+1)
设 f(x)=lnx-2(x-1)/(x+1),f(1)=0
∵ f ′(x)=(1/x)[(x-1)/(x+1)]2 ≥ 0,
∴ f(x) ≥ f(1)=0
即 lnx ≥ 2(x-1)/(x+1)
(a-b)/(lna-lnb) ≤ (a+b)/2 得证
②L(a,b)=(a-b)/(lna-lnb) ≥ √(ab)
证:令√(a/b)=x ≥1,变形得:x-1/x ≥ 2lnx
设 f(x)=x-1/x-2lnx,f(1)=0
∵ f ′(x)=(1-1/x)2 ≥ 0,
∴ f(x) ≥ f(1)=0
即 x-1/x ≥ 2lnx
(a-b)/(lna-lnb) ≥ √(ab) 得证