f(x)=x(ax^2-a-lnx)
x属于 (0,\infty)
等价于g(x)=ax^2-a-lnx>=0
g(1)=0
g'(x)=2ax-1/x
g'(1)=2a-1=0
a=1/2

h(x)=2f(x)=x^3-x-2xlnx
h'(x)=3x^2-1-2(lnx+1)=3x^2-2lnx-3
h''(x)=6x-2/x=(6x^2-2)/x
h''(x)>0 x属于(1/sqrt(3),\infty)
h''(x)<0 x属于(0,1/sqrt(3))
h'(1/sqrt(3))=2ln(sqrt(3))-2<0
在(1/8,1/e)瞎几把算算就好了吧

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