① L(a,b)=(a-b)/(lna-lnb) ≤ (a+b)/2 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 |
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