summed numerator and summed denominator

If $\displaystyle\frac{a_{1}}{b_{1}},\,\ldots,\,\frac{a_{n}}{b_{n}}$ are any real fractions with positive denominators and

m\;:=\;\min\left\{\frac{a_{1}}{b_{1}},\,\ldots,\,\frac{a_{n}}{b_{n}}\right\},% \quad M\;:=\;\max\left\{\frac{a_{1}}{b_{1}},\,\ldots,\,\frac{a_{n}}{b_{n}}\right\}

are the least and the greatest (http://planetmath.org/MinimalAndMaximalNumber) of the fractions, then

\displaystyle m\;\leqq\;\frac{a_{1}\!+\!\ldots\!+\!a_{n}}{b_{1}\!+\!\ldots\!+% \!b_{n}}\;\leqq\;M.

(1)

The equality signs are valid if and only if all fractions are equal; in this case one has

\frac{a_{1}}{b_{1}}\;=\;\ldots\;=\;\frac{a_{n}}{b_{n}}\;=\;\frac{a_{1}\!+\!% \ldots+\!a_{n}}{b_{1}\!+\!\ldots\!+\!b_{n}}.

Proof. Set $\displaystyle q_{1}:=\frac{a_{1}}{b_{1}}$ , …, $\displaystyle q_{n}:=\frac{a_{n}}{b_{n}}$ . Then we have $a_{1}\!+\!\ldots\!+\!a_{n}=b_{1}q_{1}\!+\!\ldots\!+\!b_{n}q_{n}$ , which apparently has the lower bound $(b_{1}\!+\cdots+\!b_{n})m$ and the upper bound $(b_{1}\!+\ldots+\!b_{n})M$ . Dividing the three last expressions by the sum $b_{1}\!+\ldots+\!b_{n}$ yields the asserted double inequality (1).

Remark. Cf. also the mediant.

Title	summed numerator and summed denominator
Canonical name	SummedNumeratorAndSummedDenominator
Date of creation	2013-10-11 15:35:42
Last modified on	2013-10-11 15:35:42
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	11
Author	pahio (2872)
Entry type	Theorem
Classification	msc 11A99
Related topic	InequalityForRealNumbers