# 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 SummedNumeratorAndSummedDenominator 2013-10-11 15:35:42 2013-10-11 15:35:42 pahio (2872) pahio (2872) 11 pahio (2872) Theorem msc 11A99 InequalityForRealNumbers