Let g(x)=f(x)-(mx-m)

After differentiation, there are:

g(x)=(x^2+(2-4m)x+4m+ 1)/4

Let g (x) >: =0, and its molecules form the root discriminant of the equation x 2+(2-4m) x+4m+1= 0 < = 0, that is

(2-4m)^2-4(4m+ 1)<； =0

The second question:

Exceeding this part is very simple,1/f (1)+1/f (2)+1/f (3) = 61/36 > 5/3, the following formula >; 0

Can't talk about the part.

1/f(4)+ 1/f(5)+...+ 1/f(n)=4[ 1/(4+ 1)^2+ 1/(5+ 1)^2+...+ 1/(n+ 1)^2]<； 4[ 1/(4*5)+ 1/(5*6)...+ 1/(n *(n+ 1))]= 4[ 1/4- 1/5+ 1/5- 1/6+...+ 1/n- 1/(n+ 1)]& lt； 4[ 1/4- 1/(n+ 1)]& lt； 1

Therefore, the original formula < 61/36+1= 97/36 <: 25/9