$f\in C[a, b] $ Then show that $\frac{1}{M}\int_{a}^{b} f(x) \mathrm{dx} +m\int_{a}^{b} \frac1{f(x)} \mathrm{dx} \ge 2\sqrt{\frac mM} (b-a) $

Question

Let $f\in C[a, b] $. Assume that $m> 0$ and $M$ are minimum and maximum values of $f$ over the interval $[a, b]$.Then show that $\frac{1}{M}\int_{a}^{b} f(x) \mathrm{dx} +m\int_{a}^{b} \frac 1{f(x)} \mathrm {dx} \ge 2\sqrt{\frac mM} (b-a) $ $\bf{Try} :$ If we can say $\int_{a}^{b} \frac {1} {f(x)} \mathrm dx =\frac {1}{\int_{a}^{b} f(x)} \mathrm dx$ $\bf (\text{I don't know the proof of this claim} )$ then taking $\int_{a} ^{b} f(x) \mathrm dx =y$ we can apply A.M.-G.M. inequality to get
$\frac yM +\frac my \geq 2\sqrt{\frac mM} $.
$\bf \text{But how the factor (b-a) will come in the right hand side?} $ I want to know the answer of both the questions. A alternative way of solving the problem will be interesting to learn if exists.

Answer 1

$\int_{a}^{b} \frac {1} {f(x)} \mathrm dx =\frac {1}{\int_{a}^{b} f(x)} \mathrm dx$ does not hold in general – try it with $f(x) = x$ on $[a,b]=[1, 2]$.

But you are on the right track with the AM-GM inequality. If you apply it to the integrands, not to the integrals, then everything figures out nicely: $$ \frac{1}{M}\int_{a}^{b} \, f(x) dx +m\int_{a}^{b} \frac 1{f(x)} \, dx = \int_a^b \left( \frac{f(x)}{M}+ \frac{m}{f(x)}\right) \, dx \\ \ge \int_a^b 2 \sqrt \frac mM \, dx = 2 \sqrt \frac mM (b-a) \, . $$

This inequality holds as long as $m$, $M$ and $f(x)$ are strictly positive . The fact that $m$ is the minimum and $M$ the maximum of $f$ is not used.

Equality holds if $\frac{f(x)}{M}= \frac{m}{f(x)}$ on $[a, b]$, that is if $f$ is constant.

Answer 2

Applying AM-GM once we get $$\frac{1}{M}\int_{a}^{b} f(x) \mathrm{d}x +m\int_{a}^{b} \frac1{f(x)} \mathrm{d}x \ge 2\sqrt{\frac{m}{M}}\sqrt{\underbrace{\int_{a}^{b} f(x) \mathrm{d}x\int_{a}^{b} \frac1{f(x)} \mathrm{d}x}_{\text{we need } (b-a)^2 \text{ here}}}$$ So, we need to show that $$\int_{a}^{b} f(x) \mathrm{d}x\int_{a}^{b} \frac1{f(x)} \mathrm{d}x \ge (b-a)^2$$ which is obvious according to the Cauchy-Bunyakovsky-Schwarz inequality: $$(b-a)^2 = \left(\int_a^b 1~ \mathrm dx\right)^2 = \left(\int_a^b \sqrt{f(x)}\sqrt{\frac{1}{f(x)}}\mathrm dx\right)^2 \le \int_{a}^{b} f(x) \mathrm{d}x\int_{a}^{b} \frac1{f(x)} \mathrm{d}x$$