If $f ∈ L^1((0, 1))$ is non-negative and $\int_0^1 f=1$, then $\int_0^1 f^{-1}\ge 1$

Question

Assume that $f ∈ L^1((0, 1))$ is a non-negative real valued function satisfying $$\int_{0}^{1}f(x) dx = 1 $$ Show that $$\int_{0}^{1} \frac{1}{f(x)} dx≥ 1 $$

if we can show $ \frac{1}{f(x)}$ > 1 a.e. then are not we done? I like to begin by contradiction, assume there is a positive measurable subset $A$ of $(0,1)$ such that $ \frac{1}{f(x)}<1$ on $A$. am I in right track?

Answer 1

For a self-contained proof, observe that $t+\dfrac1t\ge 2$ for all $t>0$ and conclude that $$ \int_0^1 \frac{1}{f(x)}\,dx \ge \int_0^1 (2-f(x))\,dx = 2-1=1 $$

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Jensen's inequality, used with the convex function $\phi(x) = 1/x$, also yields the claim; in a way, the above argument is an adaptation of its proof.

Cauchy-Schwarz also works, as noted in comments.