Im looking at the solution of some exercise problem. It uses the following inequality
$$\frac{1}{\sigma} \int_0^{\theta-\theta_0} \left(\exp(x/\sigma) \cdot A - \frac{1}{2\sigma}\right)^2 \,dx \leq \left(A - \frac{1}{2\sigma}\right)^2 \left(\exp(\frac{\theta-\theta_0}{\sigma}) - 1\right)$$
I struggle to derive this inequality. We have $\sigma > 0, \theta - \theta_0 > 0$. What is happening here? How can I approach this problem?
This doesn't look right. Set $\sigma=A=1$ and to make it simpler rename $t=\theta-\theta_{0}$.
So the inequality becomes $$ \int_{0}^{t}\bigg(e^x-\frac{1}{2}\bigg)^{2} dx\leq \frac{1}{4}(e^{t}-1) $$
The function in the integral is $e^{2x}-e^{x}+\frac{1}{4}$ so an antiderivate is $\frac{1}{2}e^{2x}-e^{x}+\frac{1}{4}x$. Then FTC makes the previous inequality into $$ \frac{1}{2}e^{2t}-e^{t}+\frac{1}{4}t +\frac{1}{2}\leq \frac{1}{4}(e^{t}-1) $$
Multiply through by $4$ to make it nicer, and move some things around: $$ 2e^{2t}+t+3\leq 5e^{t} $$
But this false for any $t>0$ (at least).