How do I show the following integral inequality?

Question

I have the next exercise.

Let $f:[0,1]\to\mathbb{R}$ be a continuous function. Show that $\int_{0}^{1}f^2(t)dt\geq(\int_{0}^{1}f(t)dt)^2$.

I know it can be done using Cauchy-Bunyakovsky-Schwarz inequality, but can is there any other way to solve it?

Answer 1

Hint. Note that for any real $x$, $$0\leq \int_{0}^{1}(x+f(t))^2dt=x^2+2x\int_{0}^{1}f(t)dt+\int_{0}^{1}f^2(t)dt$$ Now let $x=-\int_{0}^{1}f(t)dt$.