Let f' (x) =1-1/x = 0 = = >; x= 1

f''(x)= 1/x^2>； 0

∴f(x) takes the minimum value 1 at x= 1, and decreases monotonously when x∈(0, 1); X∈( 1, e] monotonically increases;

(2) Analysis: ∫g(x)= lnx/x x ∈( 0, e)

g'(x)=( 1-lnx)/x^2>； =0

∴g(x) monotonically increases on (0, e);

(3) Proof: ∫g(x)= lnx/x

Let g' (x) = (1-lnx)/x2 = 0 = = >; x=e

When x∈(0, e), g'(x) >; 0; When x∈(e, +∞), g' (x) < 0;

∴g(x) the maximum value when x=e g (e) =1/e;

∫ 1/e+ 1/2 & lt； 1

(0, e), f(x) ∴ > g(x)+ 1/2。