I have this exercise : let $f(X): [0,1] \rightarrow \mathbb{R}$ be a measurable function such that $ 0< m \le f(x) \le M < \infty $ a.e. Prove that
$$\big( \int_0^1 f(x) dx \Big) \big( \int_0^1 \frac{1}{f(x)} dx \Big) \le \frac{(M+m)^2}{4mM} .$$
My attempt: I tried to divide the interval $[0,1]$ in thwo sets $A=\{x\in [0,1] | f(x) \le \sqrt{Mm}\}$ and $B=[0,1] \setminus A$. Called $x$ the measure of $B$, i get that the first integral is less than $$xM+(1-x)\sqrt{Mm}$$ and the second integral is less than $$\frac{x}{\sqrt{Mm}}+\frac{(1-x)}{m}.$$ Taking the product of these two quantities and maximize for $x \in [0,1]$ I get a value (always when $x=1/2$ ) of $$ \frac{(\sqrt{M}+\sqrt{m})^2}{4m}.$$ This is similar but not exactly the bound the exercise asks to find.