# Proofs of Schweitzer's Inequality

### Preliminaries

P. Schweitzer proved the following inequality in a 1914 paper:

We already have two proofs on earlier pages:

The latter is an extract from D. A. Mitrinovic's book that was brought to my attention by N. N. Taleb. The book contains a wealth of information. Below I borrow from the book two inequalities (Diaz-Metcalf and Grüss) with proofs that are used to deliver two additional proofs of Schweitzer's inequality. The derivation of Schweitzer's inequality from Grüss' inequality is by N. N. Taleb.

**Reference**

D. S. Mitrinovic, Analytic Inequalities, Springer 1970

### Diaz-Metcalf Inequality (1963)

If $a_k\,(\ne 0)\,$ and $b_k,\,$ $k\in\overline{1,n},\,$ are real numbers and if

(1)

$\displaystyle m\le\frac{b_k}{a_k}\le M,\,\text{for}\,k\in\overline{1,n},$

then

(2)

$\displaystyle \sum_{k=1}^{n}b_k^2+mM\sum_{k=1}^{n}a_k^2\le (m+M)\sum_{k=1}^{n}a_kb_k.$

Equality holds iff in each of the $n\,$ inequalities in (1) at least one equality sign holds, i.e., $b_k=ma_k\,$ or $b_k=Ma_k\,$ (where the equation may vary with $k).$

**Proof**

By virtue of (1), $\displaystyle 0\le\left(\frac{b_k}{a_k}-m\right)\left(M-\frac{b_k}{a_k}\right)\,$ is obvious. Summing up

(3)

$\displaystyle 0\le\sum_{k=1}^{n}\left(\frac{b_k}{a_k}-m\right)\left(M-\frac{b_k}{a_k}\right),$

i.e.,

$\displaystyle 0\le -\sum_{k=1}^{n}[b_k^2-(m+M)a_kb_k+mMa_k^2]$

which is (2).

In inequality (3), or equivalently, in inequality (2), equality holds iff each of the summands $(b_k-ma_k)(Ma_k-b_k)\,$ is equal to zero, which proves the statement.

### Grüss' Inequality (1935)

Let $f\,$ and $g\,$ be two functions defined and integrable over $(a,b).\,$ Let

$\phi\le f(x)\le\Phi,\,\gamma\le g(x)\le \Gamma$

for all $x\in (a,b),\,$ where $\phi,\Phi,\gamma,\Gamma\,$ are fixed real constants. Then

(4)

$\displaystyle \left|\frac{1}{b-a}\int_{a}^{b}f(x)g(x)dx-\frac{1}{(b-a)^2}\int_{a}^{b}f(x)dx\int_{a}^{b}g(x)dx\right|\\ \qquad\qquad\le\frac{1}{4}(\Phi-\phi)(\Gamma-\gamma).$

**Proof**

First, by making the substitution $\displaystyle x=\frac{t-a}{b-a}\,$ the problem is reduced to the special case $a=0,\,b=1.\,$ Denote $F=\displaystyle \int_{0}^{1}f(x)dx,\,$ $G=\displaystyle \int_{0}^{1}g(x)dx,\,$ and consider

$\displaystyle D(f,g)=\int_{0}^{1}f(x)g(x)dx-FG.$

Then (4) reads

(5)

$\displaystyle D(f,g)\le\frac{1}{4}(\Phi-\phi)(\Gamma-\gamma).$

Note that

(6)

$D(f,f)=\displaystyle \int_0^1f^2(x)dx-\left(\int_0^1f(x)dx\right)^2\ge 0,\,$

by the *Cauchy-Schwarz inequality*. On the other hand,

$\displaystyle D(f,f)=(\Phi-F)(F-\phi)-\int_0^1[\Phi-f(x)][f(x)-\phi]dx,$

which implies

(7)

$\displaystyle D(f,f)\le (\Phi-F)(F-\phi).$

One can easily verify that

$\displaystyle D(f,g)\le \int_0^1(f(x)-F)(g(x)-G)dx.$

Using the *Cauchy-Schwarz inequality* inequality again, we get

$\displaystyle \begin{align} D(f,g)^2 &\le \int_0^1[f(x)-F]^2dx\int_0^1[g(x)-G]^2dx\\ &=D(f,f)D(g,g). \end{align}$

According to (6) and (7), we infer that

(8)

$\displaystyle D(f,g)^2\le (\Phi-F)(F-\phi)(\Gamma-G)(G-\gamma).$

Since

$4(\Phi-F)(F-\phi)\le (\Phi-\phi)^2,\\ 4(\Gamma-G)(G-\gamma)\le (\Gamma-\gamma)^2,$

we conclude that (8) implies (5).

Taking $f(x)=g(x)=sgn (2x-1),\,$ the constant $\displaystyle \frac{1}{4}\,$ in (5) is seen to be the best possible.

### Schweitzer's from Diaz-Metcalf

Letting $b_k^2=x_k,\,$ $\displaystyle a_k^2=\frac{1}{x_k}\,$ and assuming $m\le x_k\le M,\,$ so $\displaystyle \sqrt{m}\le b_k=\frac{1}{a_k}\le \sqrt{M},\,$ such that $\displaystyle m\le\frac{b_k}{a_k}\le M.\,$ The Diaz-Metcalf inequality then takes the form

$\displaystyle \sum_{k=1}^{n}x_k+mM\sum_{k=1}^{n}\frac{1}{x_k}\le (m+M)n,$

from which, employing the AM-GM inequality and subsequently squaring, we get

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

Omitting the middle part gives Schweitzer's inequality.

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

### Schweitzer's from Grüss

We shall prove Schweitzer's inequality in integral form. Let in Grüss' inequality, $m\le f(x)\le M\,$ and $g(x)=\displaystyle \frac{1}{f(x)}.\,$ The inequality becomes

$\displaystyle \left|\frac{1}{b-a}\int_a^bdx-\frac{1}{(b-a)^2}\int_a^nf(x)dx\int_a^b\frac{dx}{f(x)}\right|\\ \qquad\le\displaystyle \frac{1}{4}(M-m)\left(\frac{1}{m}-\frac{1}{M}\right)=\frac{(M-m)^2}{4mM}.$

This in particular means that

$\displaystyle \frac{1}{(b-a)^2}\int_a^nf(x)dx\int_a^b\frac{dx}{f(x)}\le\frac{(M-m)^2}{4mM}+1=\frac{(M+m)^2}{4mM}. $

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