解:(1)由f(x)=ln x+
,得f′(x)=
.
①證明:因為x>1時,h(x)=
>0,所以函數(shù)f(x)具有性質(zhì)P(b).
②當b≤2時,由x>1得x
2-bx+1≥x
2-2x+1=(x-1)
2>0,
所以f′(x)>0.從而函數(shù)f(x)在區(qū)間(1,+∞)上單調(diào)遞增.
當b>2時,令x
2-bx+1=0得
x
1=
,x
2=
.
因為x
1=
=
<
<1,
x
2=
>1,
所以當x∈(1,x
2)時,f′(x)<0;當x∈(x
2,+∞)時,f′(x)>0;當x=x
2時,f′(x)=0.從而函數(shù)f(x)在區(qū)間(1,x
2)上單調(diào)遞減,在區(qū)間(x
2,+∞)上單調(diào)遞增.
綜上所述,當b≤2時,函數(shù)f(x)的單調(diào)增區(qū)間為(1,+∞);
當b>2時,函數(shù)f(x)的單調(diào)減區(qū)間為(1,
),單調(diào)增區(qū)間為(
,+∞).
(2)由題設(shè)知,g(x)的導(dǎo)函數(shù)
g′(x)=h(x)(x
2-2x+1),
其中函數(shù)h(x)>0對于任意的x∈(1,+∞)都成立,
所以當x>1時,g′(x)=h(x)(x-1)
2>0,
從而g(x)在區(qū)間(1,+∞)上單調(diào)遞增.
①當m∈(0,1)時,
有α=mx
1+(1-m)x
2>mx
1+(1-m)x
1=x
1,
α<mx
2+(1-m)x
2=x
2,即α∈(x
1,x
2),
同理可得β∈(x
1,x
2).
所以由g(x)的單調(diào)性知g(α),g(β)∈(g(x
1),g(x
2)),從而有|g(α)-g(β)|<|g(x
1)-g(x
2)|,符合題意.
②當m≤0時,α=mx
1+(1-m)x
2≥mx
2+(1-m)x
2=x
2,β=(1-m)x
1+mx
2≤(1-m)x
1+mx
1=x
1,于是由α>1,β>1及g(x)的單調(diào)性知g(β)≤g(x
1)<g(x
2)≤g(α),
所以|g(α)-g(β)|≥|g(x
1)-g(x
2)|,與題意不符.
③當m≥1時,同理可得α≤x
1,β≥x
2,
進而得|g(α)-g(β)|≥|g(x
1)-g(x
2)|,與題意不符.
綜上所述,所求的m的取值范圍為(0,1).