I've an exercise which is the following:
Gronwall’s Inequality
Let $A > 0, B \geq 0$. Let $(\epsilon_j)_{j \in \mathbb{N}}$ be a sequence of real numbers with $$|\epsilon_{j+1}| \leq (1 + A) | \epsilon_j| + B$$ for $j=0,1 ...$.
Prove by induction that, for $A > B$ it holds that $$|\epsilon_j| \leq |\epsilon_0| \exp(jA) + \frac{B}{A}(\exp(jA) - 1)$$ for $j = 0,1..$
So, here's my proof attempt, which I think is correct, but I'm not $100\%$ sure.
Base case
$j = 0$
$$|\epsilon_0| \leq |\epsilon_0| e^{0A} + \frac{B}{A}(e^{0A} - 1) = |\epsilon_0|$$
Induction hypothesis
Lets assume that $$|\epsilon_j| \leq |\epsilon_j| e^{jA} + \frac{B}{A}(e^{jA} - 1)$$ is true for an arbitrary $j$.
Induction step
$$|\epsilon_{j+1}| \leq (1 + A) |\epsilon_j| + B$$
Now, I'm going to use the induction hypothesis
$$|\epsilon_{j+1}| \leq (1 + A) (|\epsilon_j| * e^{jA} + \frac{B}{A}(e^{jA} - 1)) + B$$
Now I use the fact that $(1 + A) \leq e^{A}$.
$$|\epsilon_{j+1}| \leq e^{A}|\epsilon_j| e^{jA} + \frac{B}{A}(e^{jA} - 1)e^{A} + B$$
$$|\epsilon_{j+1}| \leq e^{A(j + 1)}|\epsilon_j| + \frac{B}{A}(e^{jA}e^{A} - e^{A}) + B$$
$$|\epsilon_{j+1}| \leq e^{A(j + 1)}|\epsilon_j| + \frac{B}{A}(e^{A(j + 1)} - e^{A}) + B$$
$$|\epsilon_{j+1}| \leq e^{A(j + 1)}|\epsilon_j| + \frac{B}{A}((e^{A(j + 1)} - e^{A}) + A)$$
Now I use again this fact $(1 + A) \leq e^{A}$, which is equivalent to $A - e^{A}\leq -1$
so we have
$$|\epsilon_{j+1}| \leq e^{A(j + 1)}|\epsilon_j| + \frac{B}{A}(e^{A(j + 1)} - e^{A} + A)$$
$$|\epsilon_{j+1}| \leq e^{A(j + 1)}|\epsilon_j| + \frac{B}{A}(e^{A(j + 1)} - 1)$$
What I'm not sure about is the last step (but this doesn't mean there aren't other errors before). My idea is that if we sum a greater number to any other number the result will be a greater number than summing a smaller number to the corresponding any number.