Applying Landau's Inequality to Prove Bounds on f'(x)

[tomboi03]

a. Suppose f is twice differentiable on (0, infinity). Suppose that |f(x)|< or equal A0 for all x>0 and that the second derivative satisfies |f''(x)|< or equal A2 for all x>0.
Prove that for all x>0 and all h>0
|f'(x)| < or equal 2A0/h + A2h/2
This is sometimes called Landau's inequality.

b. Use part a to show that for all x>0
|f'(x)| < or equal 2 sqrt(A0A2)


I have no idea how to go about this problem.
Should I just try to do this problem backwards by trying to find what f(x) by using the f'(x)?

Thanks

 

[HallsofIvy]

Looks to me like the mean value theorem should work. If f is twice differentiable on the positive numbers, then you can apply the mean value theorem to f' on any interval of positive numbers: (f'(x)- f'(x₀))/(x- x₀)= f"(c) where c is between x₀ and x. Taking x- x₀= h, that is f'(x)- f'(x₀)= f"(c)h.
Also use f(x)- f(x₀)= f'(d)h, the mean value theorem applied to f.

 

[tomboi03]

I'm not sure how to do this, can you elaborate how to do this esp with the mean value theorem?

Thank you