Given a number of different fractions, create a new fraction whose numerator is the sum of all those fractions' numerators and whose denominator is the sum of the denominators. Call it y. Call the smallest of the original fractions x and the largest z.
Prove that for all cases, x < y < z.
Suppose we have a
1/b
1 < a
2/b
2 < ... < a
n/b
n. These imply that a
1b
i<=a
ib
1 for all i, with equality iff i=1. Adding the inequalities yields a
1(b
1+b
2+...+b
n)<b
1(a
1+a
2+...+a
n), or a
1/b
1<(a
1+a
2+...+a
n)/(b
1+b
2+...+b
n), which proves the first inequality. The second one follows analogous.
Solution
