a,b,m∈R+,且a

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a,b,m∈R+,且a

a,b,m∈R+,且a
a,b,m∈R+,且a

a,b,m∈R+,且a
a/b-(a+m)/(b+m)
=[a(b+m)-b(a+m)]/b(b+m)
=(ab+am-ab-bm)/b(b+m)
=m(a-b)/b(b+m)
a,b,m都是正数
a

证明:
交叉相乘,原不等式等价于
b(a+m)>a(b+m)
<=>ab+bm>ab+am
<=>bm>am
<=>b>a
由条件知显然成立,得证。。

设a/b=k,则a=k*b,k<1;
(a+m)/(b+m)=(k*b+m)/(b+m)=k*(b+m/k)/(b+m),
因为k<1,所以(b+m/k)/(b+m)>1,所以k<(a+m)/(b+m),即a/b<(a+m)/(b+m)