TITLE: How to prove that $\int_{0}^{1} (f'(x))^2 dx \geq \frac{2}{e^2}$
QUESTION [3 upvotes]: 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?

REPLY [2 votes]: 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.
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