PlanetMath: summed numerator and summed denominator

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summed numerator and summed denominator

(Theorem)

If $\displaystyle\frac{a_1}{b_1},\,\ldots,\,\frac{a_n}{b_n}$ are any real fractions with positive denominators and

$\displaystyle m:= \min\left\{\frac{a_1}{b_1},\,\ldots,\,\frac{a_n}{b_n}\right\}, \quad M:= \max\left\{\frac{a_1}{b_1},\,\ldots,\,\frac{a_n}{b_n}\right\}$

are the least and the greatest 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

$\displaystyle \frac{a_1}{b_1} = \ldots = \frac{a_n}{b_n} = \frac{a_1\!+\!\ldots+\!a_n}{b_1\!+\!\ldots\!+\!b_n}.$

Proof. Set $\displaystyle q_1 := \frac{a_1}{b_1}$ , ..., $\displaystyle q_n := \frac{a_n}{b_n}$ . Then we have $a_1\!+\!\ldots\!+\!a_n = b_1q_1\!+\!\ldots\!+\!b_nq_n$ , 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).

"summed numerator and summed denominator" is owned by pahio.

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See Also: inequalities for real numbers

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Cross-references: inequality, sum, expressions, upper bound, lower bound, proof, equality, denominators, positive, fractions, real
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This is version 6 of summed numerator and summed denominator, born on 2005-07-01, modified 2007-03-14.
Object id is 7203, canonical name is SummedNumeratorAndSummedDenominator.
Accessed 1265 times total.

Classification:

AMS MSC:

11-01 (Number theory :: Instructional exposition )

Pending Errata and Addenda

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