证明:(1)令H(x)=(x+m)ln﹣2(x﹣m),x∈(m,+∞), 则H(m)=0, 要证明(x+m)ln﹣2(x﹣m)>0, 只需证H(x)=(x+m)ln﹣2(x﹣m)>H(m), ∵H′(x)=ln+﹣1, 令G(x)=ln+﹣1,G′(x)=﹣, 由G′(x)=>0得,x>m, ∴G(x)在x∈(m,+∞)单调递增, ∴G(x)>G(m)=0 H"(x)>0,H(x)在x∈(m,+∞)单调递增. H(x)>H(m)=0, ∴H(x)=(x+m)ln﹣2(x﹣m)>0, (2)不妨设0<x1<x2, 要证(x1+x2)g(x1+x2)>2, 只需证(x1+x2)[a(x1+x2)+b]>2, 只需证(x1+x2)[a+bx2﹣(a+bx1)]>2(x2﹣x1), ∵=ax1+b,=ax2+b, 即(x1+x2)ln>2(x2﹣x1)(*), 而由(1)知(*)成立. 所以(x1+x2)g(x1+x2)>2 |