f(x)=ln(x-1)+2a/x,
f'(x)=1/(x-1)-2a/x^2
=[x^2-2a(x-1)]/[(x-1)x^2]
=(x^2-2ax+2a)/[(x-1)x^2](x>1),
a0,f(x)↑；
a=2时f'(x)=(x-2)^2/[(x-1)x^2]>=0,f(x)↑；
a>2时a-√（a^2-2a)>=1,
a-1>=√(a^2-2a),显然成立.
a-√(a^2-2a)