Inequalities for combinations of $\int f $ and $\int (1/f)$ where $m\le f\le M$ on an interval

Question

Let $f\in C[a,b]$. Assume that $\min_{[a,b]}f=m>0$ and $M=\max_{[a,b]}f$.
Which one is true?

a. $$\frac{1}{M}\int_a^bf(x)dx+m\int_a^b\frac{1}{f(x)}dx\geq 2\sqrt{\frac{m}{M}}(b-a)$$

b. $$\int_a^bf(x)dx\int\frac{1}{f(x)}dx\geq (b-a)^2$$

c. $$\int f(x)dx\int_a^b\frac{1}{f(x)}dx\leq (b-a)^2$$

For the first one $$\frac{1}{M}\int _a^bf(x)+m\int _a^b\frac{1}{f(x)}dx\geq {\frac{m}{M}}(b-a)+{\frac{m}{M}}(b-a)$$ and the last two I am not getting.
How to find these?

Answer 1

For a. Note that if $x > 0$, we have ${x\over M} + {m \over x} \ge 2 \sqrt{m \over M}$ (the function is strictly convex and unbounded as $x \downarrow 0$ and $x \to \infty$, differentiate to find the minimiser).

$\int_a^b ({f(x) \over M} + {m \over f(x)})dx \ge \int_a^b2 \sqrt{m \over M} dx$.

For b. use Cauchy-Schwarz with the functions $x \mapsto \sqrt{f(x)}$ and $x \mapsto \sqrt{1 \over f(x)}$.

Let $g(x) = \sqrt{f(x)}, h(x) = \sqrt{1 \over f(x)}$. Then $(\int_a^b g(x)h(x)dx )^2 \le \int_a^b g(x)^2 dx \int_a^b h^2(x) dx$.

For c. pick any non-constant $f$ that satisfies the hypotheses.

Answer 2

In fact,we have this following stronger inequality: $$\left(\int_{a}^{b}f(x)dx\right)\cdot\left(\int_{a}^{b}\dfrac{1}{f(x)}dx\right)\le \dfrac{(m+M)^2(b-a)^2}{4Mm}$$ proof :since $ m\le f(x)\le M$ so we have $$\dfrac{(f-M)(f-m)}{f}<0\Longrightarrow f+\dfrac{Mm}{f}\le M+m$$ so $$\int_{a}^{b}f(x)dx+Mm\int_{a}^{b}\dfrac{1}{f(x)}dx\le \int_{a}^{b}(M+m)dx=(M+m)(b-a)$$ use AM-GM inequality we have $$\int_{a}^{b}f(x)dx+Mm\int_{a}^{b}\dfrac{1}{f(x)}dx\ge 2\sqrt{Mm\left(\int_{a}^{b}f(x)dx\right)\cdot\left(\int_{a}^{b}\dfrac{1}{f(x)}dx\right)}$$ so we have $$\left(\int_{a}^{b}f(x)dx\right)\cdot\left(\int_{a}^{b}\dfrac{1}{f(x)}dx\right)\le \dfrac{(m+M)^2(b-a)^2}{4Mm}$$

Other hand use Cauchy-Schwarz inequality we have $$\left(\int_{a}^{b}f(x)dx\right)\cdot\left(\int_{a}^{b}\dfrac{1}{f(x)}dx\right)\ge (b-a)^2$$