f'(x)=-(lnx+ 1)/(xlnx)^2

The intervals of simple increase (0, 1/e) and monotonic decrease (1/e, 1) are obtained.

Equation 2 (1/x) >: take the logarithm with the base of e on both sides of X A at the same time, and get:

ln(2^( 1/x))>； ln(x^a)

That is LN2/X >；; alnx

Because the value range of x is (0, 1), so

a & gtln2/(xlnx)

Therefore, a should be greater than the maximum value of ln2/(xlnx).

Because f(x)= 1/(xlnx) is obtained from monotone interval, f(x) is maximum when the maximum value is x =1/e.

So a>- El n2