Question

不等式的性质:a>0,b>0,m>0,求证：b+m/a+m>b/a

Answer 1

应改成 a>b>0,m>0,求证：(b+m)/(a+m)>b/a

(b+m)/(a+m) - b/a =a(b+m)/a(a+m) - b(a+m)/a(a+m) 通分

=[ab+am - (ab+bm)] / a(a+m) = [m(a-b)] / a(a+m) ...(*)

a>b ,m>0 所以 m(a-b)正数 ; a(a+m)正数 => (*)是正数

=> 够减 (b+m)/(a+m) > b/ a