当x2>x1>1时,证明lnx1除以lnx2小于x2除以x1 若x1>x2>0,求证（lnx1-lnx2）/(x1-x2 ...
\u8bc1\u660e\uff1ax1-x2/lnx1-lnx2\uff1cx1+x2/2\uff0c\u8c22\u8c22\uff01\u5e94\u8be5\u662f\uff1a(x1-x2)/(lnx1-lnx2)\uff1c(x1+x2)/2 \u5427\uff1f\uff08\u8981\u6709\u62ec\u53f7\u554a\uff09
\u8981\u8bc1\uff1a(x1-x2)/(lnx1-lnx2)\uff1c(x1+x2)/2
\u5373\u8bc1\uff1aln(x2/x1)>2(x2-x1)/(x1+x2)
\u5373\u8bc1\uff1aln(x2/x1)>2[(x2/x1)-1]/[1+(x2/x1)]
\u56e0\u4e3a\uff1a01
\u4e8e\u662f\u53ea\u9700\u8bc1:f(x)=lnx-2(x-1)/(x+1)>0\u5728x>1\u65f6\u6052\u6210\u7acb
\u56e0\u4e3a\uff1af\uff07(x)=1/x-4/(x+1)^2=(x-1)^2/[x(x+1)^2] >0
\u6240\u4ee5\uff1af(x)\u5728x>1\u65f6\u5355\u8c03\u9012\u589e,
\u56e0\u4e3a\uff1af(1)=0
\u6240\u4ee5\uff1a:f(x)>0\u5728x>1\u65f6\u6052\u6210\u7acb
\u5373\u8bc1(x1-x2)/(lnx1-lnx2)\uff1c(x1+x2)/2
\u672c\u9898\u4e3b\u8981\u8003\u5bdf\u7684\u662f\u4e0d\u7b49\u5f0f\u7684\u7efc\u5408\u5206\u6790\uff0c\u4ee5\u53ca\u6784\u9020\u51fd\u6570\u6765\u8bc1\u660e\u4e0d\u7b49\u5f0f\uff0c\u5c5e\u4e8e\u4e2d\u6863\u9898\u76ee\u3002
要证(lnx1)/(lnx2)<x2/x1
只需证 x1·(lnx1)<x2·(lnx2)
设f(x)=xlnx
f'(x)=lnx+1
x∈(1,+∞)时，f'(x)=lnx+1>(ln1)+1>0
得f(x)是(1,+∞)上的增函数
因x2>x1>1，得f(x1)<f(x2)
即 x1·(lnx1)<x2·(lnx2)成立。
所以x2>x1>1时，(lnx1)/(lnx2)<x2/x1
希望能帮到你！
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