So -m=- 1, the solution is m= 1.
Because f '(x)= 1
1
x
- 1=0,
So the slope of the tangent is 0,
So the tangent equation is y=- 1.
(2) Because f '(x)= 1
1
x
-m=
1-mx
x
.
① when m≤0, x∈( 1, e) and f ′ (x) > 0,
Therefore, the function f (x) monotonically increases at (1, e),
Then f (x)max=f (e)= 1-me.
② When?
1
m
≥e, that is, 0 < m ≤
1
e
X∈( 1，e)，f′(x)> 0，
Therefore, the function f (x) monotonically increases at (1, e),
Then f (x)max=f (e)= 1-me.
③ When 1
1
m
< e, i.e.
1
e
< m < 1，
The function f (x) is in (1,
1
m
) monotonically increasing, in (
1
m
E) monotonically decreasing,
Then f (x)max=f (
1
m
)=-lnm- 1。
④ When?
1
m
≤ 1, that is, when m≥ 1, x∈( 1, e), f ′ (x) < 0,
The function f (x) monotonically decreases at (1, e),
Then f (x) max = f (1) =-m.
To sum up, ① when m≤
1
e
f(x)max = 1-me；
② When?
1
e
< m < 1，f(x)max =-lnm- 1；
③ When m≥ 1, f (x) max =-m. 。
(3) let x 1>x2>0 > X2 > 0.
Because f (x 1)=f (x2)=0,
So lnx 1-mx 1=0, lnx2-mx2=0,
Lnx 1+lnx2=m(x 1+x2)，lnx 1-lnx2=m(x 1-x2)。
To prove that x 1x2 > E2,
That is, lnx 1+lnx2 > 2, that is, m (x 1+x2) > 2.
Because m=
lnx 1-lnx2
x 1-x2
So this proves that
lnx 1-lnx2
x 1-x2
>
2
x 1+x2
Namely ln
x 1
x2
>
2(x 1-x2)
x 1+x2
.
manufacture
x 1
x2
=t, then t > 1, then lnt > >
2(t- 1)
t+ 1
.
Order? (t)=lnt-
2(t- 1)
t+ 1
(t> 1)，
then what ′(t)=
1
t
-
four
(t+ 1)2
=
(t- 1)2
t(t+ 1)2
>0.
What about the function? (t) Incremental function on (1, +∞),
So what? (t)>？ (1)=0, i.e. lnt > >
2(t- 1)
t+ 1
Established.
So the original inequality holds.