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.