Let f (x) = SiNx (x ≥ 0), G (x) = ax (x ≥ 0), where a is a real number 1. If f (x) ≤ g (x) is constant, find the value range of real number a 2. When a = 1, we prove that G (x) - f (x) ≤ (1 / 6) x & # 179; (x ≥ 0)

Let f (x) = SiNx (x ≥ 0), G (x) = ax (x ≥ 0), where a is a real number 1. If f (x) ≤ g (x) is constant, find the value range of real number a 2. When a = 1, we prove that G (x) - f (x) ≤ (1 / 6) x & # 179; (x ≥ 0)

Let H (x) = f (x) - G (x) = SiNx ax (x ≥ 0),
So h '(x) = cosx-a
If a ≥ 1, H '(x) = cosx-a ≤ 0,
So h (x) = SiNx ax decreases monotonically in the interval (- ∞, 0), that is, H (x) ≤ H (0) = 0,
So SiNx ≤ ax (x ≥ 0) holds
If a < 1, there exists x0 ∈ (0, π / 2) such that cosx0 = a,
So x ∈ (0, x0), H '(x) = cosx-a > 0,
So h (x) = SiNx ax increases monotonically over the interval (0, x0),
So there exists x such that h (x) > H (0) = 0, that is, f (x) ≤ g (x) does not hold,
Therefore, a < 1 does not conform to the meaning of the subject
In conclusion, a ≥ 1
(II) from the meaning of the question, we can get: a = 1, so g (x) = x (x ≥ 0),
So (x) - G (x) = sinx-x (x ≥ 0),
So the original inequality is equivalent to sinx-x-1 / 6x ^ 3 ≤ 0 (x ≥ 0),
Let H (x) = x-sinx-1 / 6x ^ 3 (x ≥ 0), so h ′ (x) = 1-cosx-1 / 2x ^ 2
Let g (x) = 1-cosx-1 / 2x ^ 2, so G '(x) = sinx-x,
So G '(x) = sinx-x ≤ 0 (x ≥ 0),
So g (x) = 1-cosx-1 / 2x ^ 2 decreases monotonically on (0, + ∞),
Therefore, G (x) = 1-cosx-1 / 2x ^ 2 ≤ g (0) = 0,
That is, H ′ (x) = 1-cosx-1 / 2x ^ 2 ≤ 0,
So h (x) = x-sinx-1 / 6x ^ 3 (x ≥ 0) decreases monotonically,
So h (x) = x-sinx-1 / 6x ^ 3 ≤ H (0) = 0,
So x-sinx-1 / 6x ^ 3 ≤ 0 (x ≥ 0) is constant, that is x-sinx ≤ 1 / 6x ^ 3 (x ≥ 0)