Sitaru-Schweitzer Inequality

Problem

Below is a slightly modified version of Dan Sitaru's statement from his book "Math Phenomenon." The problem represents an integral analog and a generalization of the well known Schweitzer's Inequality derived next.

Sitaru-Schweitzer Inequality

Remark

Note that, by the Cauchy criterion for integrability, if function $f:\,[a,b]\rightarrow [m,M],\,$ with $m\gt 0\,$ is Riemann integrable, then the function $\displaystyle \frac{1}{f}:\,[a,b]\rightarrow \left[\frac{1}{M},\frac{1}{m}\right]\,$ is also Riemann integrable.

Solution 1

It follows from the premises that $(f(x)-m)(M-f(x))\ge 0,\,$ implying $f^2(x)+mM\le (m+M)f(x),\,$ or

$\displaystyle f(x)+\frac{mM}{f(x)}\le m+M,$

so that

$\displaystyle \int_{a}^{b}f(x)dx+mM\int_{a}^{b}\frac{dx}{f(x)}\le (m+M)\int_{a}^{b}dx=(m+M)(b-a).$

Let $\displaystyle J=mM\int_{a}^{b}\frac{dx}{f(x)}\,$ and $\displaystyle H=\int_{a}^{b}f(x)dx.\,$ We have $J+H\le(m+M)(b-a),\,$ implying $J^2+JH\le (m+M)(b-a)J,\,$ i.e.,

$\displaystyle\begin{align} JH&\le\frac{1}{4}(m+M)^2(b-a)^2-\left[\frac{1}{2}(m+M)(b-a)-J\right]^2\\ &\le\frac{1}{4}(m+M)^2(b-a)^2. \end{align}$

This is the required inequality.

Solution 2

We proceed as above to obtain

$\displaystyle \int_{a}^{b}f(x)dx+mM\int_{a}^{b}\frac{dx}{f(x)}\le (m+M)\int_{a}^{b}dx=(m+M)(b-a).$

By the AM-GM inequality,

$\displaystyle (m+M)(b-a)\ge2\sqrt{mM\left(\int_{a}^{b}f(x)dx\right)\left(\int_{a}^{b}\frac{dx}{f(x)}\right)}.$

Squaring gives the required inequality.

Schweitzer's Inequality

For $0\lt m\lt M,\,$ and $x_k\in [m,M],\,$ for $k\in\overline{1,n},$

$\displaystyle \left(\sum_{k=1}^nx_k\right)\left(\sum_{k=1}^n\frac{1}{x_k}\right)\le\frac{(m+M)^2}{4mM}n^2.$

Proof of Schweitzer's Inequality

Given a sequence $\{x_k\}\subset [m,M],\,$ we define a function piece-wise: set, for $x\in [k,k+1],\,$ $f(x)=x_k,\,$ $k\in\overline{1,n}.\,$ Then, with $a=1\,$ and $b=n+1,\,$ $f:\,[a,b]\rightarrow [m,M],\,$ and it remains to observe that,

$\displaystyle \begin{align} &b-a=n\\ &\int_{a}^{b}f(x)dx=\sum_{k=1}^nx_k\\ &\int_{a}^{b}\frac{dx}{f(x)}=\sum_{k=1}^n\frac{1}{x_k}. \end{align}$

Note that two additional proofs appear elsewhere.

Acknowledgment

The problem has been kindly posted by Dan Sitaru at the CutTheKnotMath facebook page. Solution 1 is by Ravi Prakash; Solution 2 is by Diego Alvariz.

 

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