TITLE: $m \leq f(x) \leq M$ implies $\left( \int_0^1 \frac{dx}{f(x)}\right) \left( \int_0^1 f(x)dx\right) \leq \frac{(m+M)^2}{4mM}$
QUESTION [1 upvotes]: Let $f$ be a continous function on the interval $[0, 1]$ such that $0 < m \leq f(x) \leq M$ for all $x \in [0, 1]$. Show that
\begin{equation}
    \left( \int_0^1 \frac{dx}{f(x)}dx\right) \left( \int_0^1 f(x)dx\right) \leq \frac{(m+M)^2}{4mM}
\end{equation}
I wrote this as
\begin{equation}
\int_0^1 \int_0^1 \frac{M f(y)}{m f(x)}dxdy \leq \left(\frac{1+\frac{M}{m}}{2}\right)^2
\end{equation}
because i know that $f(y)/f(x) \leq \frac{M}{m}$ and the RHS looks like AM-GM.

REPLY [5 votes]: We have
$$\left(f(x)-m \right)\left(\frac{1}{f(x)}-\frac{1}{M} \right) \ge0 \hspace{5 mm}\forall x \in(0,1)
\implies 1+\frac{m}{M} \ge \frac{m}{f(x)} +\frac{f(x)}{M} \hspace{5 mm}\forall x \in(0,1)$$
So,
$$1+\frac{m}{M} \ge \int_0^1 \frac{m}{f(x)}dx+ \int_0^1\frac{f(x)}{M} dx  \tag{1}$$
Applying AM-GM inequality to $(1)$
$$1+\frac{m}{M} \ge 2 \sqrt{\frac{m}{M}} \sqrt{\int_0^1 \frac{dx}{f(x)}\int_0^1f(x) dx} \tag{2}$$
From $(2)$, you deduce easily your inequality. Q.E.D
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