Let $f$ be a continuous function from $[0,1]$ to $\mathbb R$ and $f>0$. I want to show that $$1\le (\int_0^1f(t)dt)(\int_0^1\dfrac{1}{f(t)}dt)$$
My idea is to use the inner product $\langle f,g \rangle =\int_0^1f(t)g(t)dt$. The Cauchy Shwarz inequality gives that $$1 \le (\int_0^1f^2(t)dt)(\int_0^1 \dfrac{1}{f^2(t)} dt)$$ but how to proceed to obtain the desired inequality?