Is $f(t) = At^{-2} + Bt^{-3} \le Ct^{-3}$?

Question

I don't have much experience with estimates so I'm not sure if I a bound I have derived is correct. If I have $f(t) = At^{-2} + Bt^{-3}$ on the interval $[\epsilon, 1]$ can I say that

$$ f(t) \le Ct^{-3}. $$

Is this true? If so, how can it be derived?

Answer 1

If you search for $C$ satisfying the inequality, then let's start from the end:

$$At^{-2}+Bt^{-3} \leq Ct^{-3}$$

Multiplying both sides by $t^3$ we get

$$At+B \leq C$$

(note that the sign was not reversed since $t$ is positive)

Now since $t\in (0,1]$ (note that $0$ has to be exclude since you take inverse originaly) then the left side is at most $\max(A+B, B)$ depending on whether $A,$ is positive or negative. Thus

$$C:=\max(A+B, B)$$

is a good enough choice.