Let y = f (x) (x ∈ R and X ≠ 0) be f (XY) = f (x) + F (y) for any nonzero real number x, y (1) Verification: F (1) = f (_ 1) And f (1 / x) = - f (x) (x ≠ 0) (2) judge the parity of F (x) (3) if f (x) monotonically increases on (0, + ∞), solve the inequality f (1 / x) - f (2x-1) ≥ 0

Let y = f (x) (x ∈ R and X ≠ 0) be f (XY) = f (x) + F (y) for any nonzero real number x, y (1) Verification: F (1) = f (_ 1) And f (1 / x) = - f (x) (x ≠ 0) (2) judge the parity of F (x) (3) if f (x) monotonically increases on (0, + ∞), solve the inequality f (1 / x) - f (2x-1) ≥ 0

(1) Let (x = f (x) (x = f (x) (x) (x (x) (x) (x) (x) (x) (x) (x) (x = f (x) (x) (x (x) (x) (x) (x) (x) (x) (x = f (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x = f (XY) = f (x (x) + F (f (f (f (f (y) + F (x (x (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x) (x \ \\\\\\\\\theoremtheorem = 0) (x) (x) (x = 0)) 0) (2) f (1) = f (- 1) = 0 is even function (3) ∵ is even function increasing on (0 + ∞), When x > 0, the solution of F (1 / x) - f (2x-1) ≥ 0  1 / X ≥ 2x-1 is x ∈ (0.1] or [- 1 / 2 - ∞)] x ∈ (0.1] when x