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Let $a<b\in\mathbb{R}$ and $f:[a,b]\to (0,\infty)$ be continuous.

Show the inequality

$$\left(\int_a^b f(x)\,dx\right)\left(\int_a^b \frac{1}{f(y)}\,dy\right)\geq (b-a)^2$$

As a hint is given to first prove that $\frac{f(x)}{f(y)}+\frac{f(y)}{f(x)}\geq 2$. This is easy.

I wonder if this "hint" is actually needed. I would have argued that $g:[a,b]^2\to (0,\infty)$, $(x,y)\mapsto \frac{f(x)}{f(y)}$ is a continuous map on a compact set, hence takes it minimum $m$.

An easy calculation then shows

$$\iint_{[a,b]^2} \frac{f(x)}{f(y)}d^2(x,y)\geq (b-a)^2$$

by estimating below with the minimum $m$, and calculating the double integral.

Every comment is helpful. Thanks in advance.

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  • $\begingroup$ The only obvious estimate I see is $\int_{[a,b]^2}\frac{f(x)}{f(y)}\,d^2(x,y)=\int_{[a,b]}g(x,y)\,d^2(x,y)\geq m\cdot (b-a)^2$. How do you conclude? $\endgroup$
    – peek-a-boo
    Dec 30, 2021 at 21:26
  • $\begingroup$ @peek-a-boo I messed up the easy calculation. LOL $\endgroup$
    – Cornman
    Dec 30, 2021 at 21:39

2 Answers 2

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From the hint: apply $\displaystyle \int_{a}^b dx$ to both sides yields: $\dfrac{1}{f(y)}\displaystyle \int_{a}^bf(x)dx+f(y)\displaystyle \int_{a}^b \dfrac{1}{f(x)}dx =\displaystyle \int_{a}^b \left(\dfrac{f(x)}{f(y)}+\dfrac{f(y)}{f(x)}\right) dx \ge \displaystyle \int_{a}^b 2dx = 2(b-a)$. Apply again $\displaystyle \int_{a}^b dy$ to both sides of this new inequality yields: $2\displaystyle \int_{a}^bf(x)dx\displaystyle \int_{a}^b \dfrac{1}{f(y)}dy \ge 2(b-a)^2$. Upon canceling out the factor of $2$ both sides leads to the desire inequality.

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We have the following : \begin{aligned}\left(\int_{a}^{b}{f\left(x\right)\mathrm{d}x}\right)\left(\int_{a}^{b}{\frac{\mathrm{d}y}{f\left(y\right)}}\right)&=\int_{a}^{b}{\int_{a}^{b}{\frac{f\left(x\right)}{f\left(y\right)}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=\int_{a}^{b}{\int_{a}^{x}{\frac{f\left(x\right)}{f\left(y\right)}\,\mathrm{d}y}\,\mathrm{d}x}+\int_{a}^{b}{\int_{x}^{b}{\frac{f\left(x\right)}{f\left(y\right)}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=\int_{a}^{b}{\int_{a}^{x}{\frac{f\left(x\right)}{f\left(y\right)}\,\mathrm{d}y}\,\mathrm{d}x}+\int_{a}^{b}{\int_{a}^{y}{\frac{f\left(x\right)}{f\left(y\right)}\,\mathrm{d}x}\,\mathrm{d}y}\\ &=\int_{a}^{b}{\int_{a}^{x}{\frac{f\left(x\right)}{f\left(y\right)}\,\mathrm{d}y}\,\mathrm{d}x}+\int_{a}^{b}{\int_{a}^{x}{\frac{f\left(y\right)}{f\left(x\right)}\,\mathrm{d}y}\,\mathrm{d}x}\\ \left(\int_{a}^{b}{f\left(x\right)\mathrm{d}x}\right)\left(\int_{a}^{b}{\frac{\mathrm{d}y}{f\left(y\right)}}\right)&=\int_{a}^{b}{\int_{a}^{x}{\left(\frac{f\left(x\right)}{f\left(y\right)}+\frac{f\left(y\right)}{f\left(x\right)}\right)\mathrm{d}y}\,\mathrm{d}x}\end{aligned}

In the third line we've exchanged the integral signs in the second term, an easy way to show that is by considering a function $ f:\mathbb{R}^{2}\rightarrow\mathbb{R} $ defined as follows : $ f:\left(x,y\right)\mapsto\left\lbrace\begin{matrix} 0,\ \ \ \ \ \ \ \ \text{If }\ a\leq y< x\leq b\\ \frac{f\left(x\right)}{f\left(y\right)},\ \ \ \text{If }\ a\leq x\leq y\leq b\end{matrix}\right. $, then : $$ \int_{a}^{b}{\int_{x}^{b}{\frac{f\left(x\right)}{f\left(y\right)}\,\mathrm{d}y}\,\mathrm{d}x}=\int_{a}^{b}{\int_{a}^{b}{f\left(x,y\right)\mathrm{d}y}\,\mathrm{d}x}=\int_{a}^{b}{\int_{a}^{b}{f\left(x,y\right)\mathrm{d}x}\,\mathrm{d}y}=\int_{a}^{b}{\int_{a}^{y}{\frac{f\left(x\right)}{f\left(y\right)}\,\mathrm{d}x}\,\mathrm{d}y} $$

And in the fourth line we substituted $ x\leftrightarrow y $ in the second term, because $ x $ and $ y $ are mute variables.

Now : \begin{aligned} \left(\int_{a}^{b}{f\left(x\right)\mathrm{d}x}\right)\left(\int_{a}^{b}{\frac{\mathrm{d}y}{f\left(y\right)}}\right)&=\int_{a}^{b}{\int_{a}^{x}{\left(\frac{f\left(x\right)}{f\left(y\right)}+\frac{f\left(y\right)}{f\left(x\right)}\right)\mathrm{d}y}\,\mathrm{d}x}\\ &\geq 2\int_{a}^{b}{\left(x-a\right)\mathrm{d}x}=\left(b-a\right)^{2} \end{aligned}

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