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 The Sum of the Parts Is Less Than And More Than the Parts (Posted on 2004-10-15)
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.

 No Solution Yet Submitted by Victor Zapana Rating: 2.3333 (3 votes)

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 Solution | Comment 1 of 11
Suppose we have a1/b1 < a2/b2 < ... < an/bn.  These imply that a1bi<=aib1 for all i, with equality iff i=1.  Adding the inequalities yields a1(b1+b2+...+bn)<b1(a1+a2+...+an), or a1/b1<(a1+a2+...+an)/(b1+b2+...+bn), which proves the first inequality.  The second one follows analogous.
 Posted by David Shin on 2004-10-15 17:07:22

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