The function f (x) = x-1-alnx (a ∈ R) is known. It is proved that f (x) ≥ 0 is constant if and only if a = 1 ② Necessity F '(x) = 1-ax = x-ax, where x > 0 (i) When a ≤ 0, f '(x) > 0 is constant, so f (x) is an increasing function on (0, + ∞) And f (1) = 0, so when x ∈ (0,1), f (x) < 0, which is contrary to f (x) ≥ 0 A ≤ 0 does not satisfy the problem (II) when a > 0, ∵ x > A, f '(x) > 0, so f (x) is an increasing function on (a, + ∞); When 0 ＜ x ＜ a, f '(x) ＜ 0, so the function f (x) is a decreasing function on (0, a); ∴f（x）≥f（a）=a-a-alna ∵ f (1) = 0, so when a ≠ 1, f (a) < f (1) = 0, which is in contradiction with F (x) ≥ 0 ∴a=1 In the process of proving the necessity above, what is the meaning of "∵ f (1) = 0, so when a ≠ 1, f (a) < f (1) = 0, which is in contradiction with F (x) ≥ 0?" why is there f (a) < f (1) when a ≠ 1?

The function f (x) = x-1-alnx (a ∈ R) is known. It is proved that f (x) ≥ 0 is constant if and only if a = 1 ② Necessity F '(x) = 1-ax = x-ax, where x > 0 (i) When a ≤ 0, f '(x) > 0 is constant, so f (x) is an increasing function on (0, + ∞) And f (1) = 0, so when x ∈ (0,1), f (x) < 0, which is contrary to f (x) ≥ 0 A ≤ 0 does not satisfy the problem (II) when a > 0, ∵ x > A, f '(x) > 0, so f (x) is an increasing function on (a, + ∞); When 0 ＜ x ＜ a, f '(x) ＜ 0, so the function f (x) is a decreasing function on (0, a); ∴f（x）≥f（a）=a-a-alna ∵ f (1) = 0, so when a ≠ 1, f (a) < f (1) = 0, which is in contradiction with F (x) ≥ 0 ∴a=1 In the process of proving the necessity above, what is the meaning of "∵ f (1) = 0, so when a ≠ 1, f (a) < f (1) = 0, which is in contradiction with F (x) ≥ 0?" why is there f (a) < f (1) when a ≠ 1?

When a > 0, ∵ x > A, f '(x) > 0, so the function f (x) is an increasing function on (a, ∞);
When 0 ＜ x ＜ a, f '(x) ＜ 0, so the function f (x) is a decreasing function on (0, a);
∴f（x）≥f（a）=a-a-alna
From the derivative we know that f (a) is the minimum