How to prove that $\int_{0}^{1} (f'(x))^2 dx \geq \frac{2}{e^2}$

Question

Here is the question I am trying to solve:

Suppose $f$ is a real, continuously differentiable function on $[0,1], f(0) = f(1) = 0,$ and $$\int_{0}^{1} f(x)e^x dx = 1.$$ Prove that $(a) \int_{0}^{1} f'(x) e^x dx = -1,$ and

$(b) \int_{0}^{1} (f'(x))^2 dx \geq \frac{2}{e^2}.$

My question is:

I know how to prove $(a)$ where I used integration by parts but I do not know how to solve $(b),$ should I use the mean value theorem or intermediate value theorem?

Can anyone help me in solving letter $(b)$ please?

Answer 1

Using Cauchy-Schwartz inequality, you have

$$\begin{aligned} 1 &= \left\lvert \int_{0}^{1} f^\prime(x) e^x dx \right\rvert^2\\ &\le \int_{0}^{1} (f^\prime(x))^2 dx \int_{0}^{1} e^{2x} dx\\ &= \left(\int_{0}^{1} (f^\prime(x))^2 dx\right) \left(\frac{1}{2}(e^2-1)\right)\\ &\le \left(\int_{0}^{1} (f^\prime(x))^2 dx\right) \frac{e^2}{2} \end{aligned}$$

leading to the desired result.