If $f:[a,b]\rightarrow (0,+\infty)$ be a continuous function, then
$$\int _a^b f(x)dx . \int _a^b \dfrac{1}{f(x)}dx \geq (b-a)^2$$
By usual inner product $\langle f,g \rangle = \int _a^b f(x)g(x)dx$, we get the following inequality
$$\int _a^b f^2(x)dx . \int _a^b \dfrac{1}{f^2(x)}dx \geq (b-a)^2$$
How to prove the first one?