Suppose $\int_{0}^{a}f(x)dx = 1$. Is it true that for every $b>0$ we have $\int_{0}^{a}\frac{e^{bx}}{b}\left(f(x)\right)^2dx \ge 1 \; ? $

Question

Suppose $a>0$ is such that $$ \int_{0}^{a}f(x)dx = 1. $$

Is it true that for every $b>0$ we have $$ \int_{0}^{a}\frac{e^{bx}}{b}\left(f(x)\right)^2dx \ge 1 \; ? $$

Answer 1

By Cauchy-Schwarz inequality, $$ \left(\int_0^a e^{bx}f^2(x)dx\right)\left(\int_0^a e^{-bx}dx\right)\ge\left(\int_0^a f(x)dx\right)^2=1. $$

So $$ \int_0^a e^{bx}f^2(x)dx\ge\frac{1}{\int_0^a e^{-bx}dx}=\frac{b}{1-e^{-ab}}>b. $$