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。