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Algebra Difficulty 6.8 National Olympiad Prove it Estonia

Anna, Berta and Carol make fruit drinks from syrup. Anna makes aa litres of drink by mixing water and syrup in the proportion of a:1a : 1. Berta makes bb litres of drink by mixing water and syrup in the proportion of b:1b : 1. Carol makes cc litres of drink by mixing water and syrup in the proportion of c:1c : 1. (It is not known if aa, bb and cc are integers.) They make 66 litres of drink in total. Prove that they use at most 22 litres of syrup.

Solutions — 3

Solution 1

The percentage of syrup in Anna's drink is 1a+1\frac{1}{a+1}, thus it contains aa+1\frac{a}{a+1} litres of syrup. Similarly, Berta uses bb+1\frac{b}{b+1} litres and Carol uses cc+1\frac{c}{c+1} litres of syrup. Hence we have to prove that a+b+c=6a + b + c = 6 implies
aa+1+bb+1+cc+1≤2. \frac{a}{a+1} + \frac{b}{b+1} + \frac{c}{c+1} \le 2.
As aa+1=1−1a+1\frac{a}{a+1} = 1 - \frac{1}{a+1} and similarly bb+1=1−1b+1\frac{b}{b+1} = 1 - \frac{1}{b+1} and cc+1=1−1c+1\frac{c}{c+1} = 1 - \frac{1}{c+1}, this is equivalent to the inequality
1a+1+1b+1+1c+1≥1.(12) \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} \ge 1. \quad (12)
Multiplying both sides by 33 leads to the equivalent inequality
3a+1+3b+1+3c+1≥3. \frac{3}{a+1} + \frac{3}{b+1} + \frac{3}{c+1} \ge 3.
Adding 33 to both sides and applying 3=6+33=a+b+c+33=a+13+b+13+c+133 = \frac{6+3}{3} = \frac{a+b+c+3}{3} = \frac{a+1}{3} + \frac{b+1}{3} + \frac{c+1}{3} in the left hand side, we obtain the equivalent inequality
3a+1+a+13+3b+1+b+13+3c+1+c+13≥6.(13) \frac{3}{a+1} + \frac{a+1}{3} + \frac{3}{b+1} + \frac{b+1}{3} + \frac{3}{c+1} + \frac{c+1}{3} \ge 6. \quad (13)
The sum of every positive real number and its reciprocal is at least 22. Hence the inequality (13) holds for every aa, bb and cc.

Solution 2

As in Solution 1, we reduce the problem to the inequality (12). After converting the fractions to a common denominator, removing parentheses and collecting similar terms, it suffices to show that
ab+bc+ca+2(a+b+c)+3abc+ab+bc+ca+a+b+c+1≥1, \frac{ab + bc + ca + 2(a + b + c) + 3}{abc + ab + bc + ca + a + b + c + 1} \ge 1,
or equivalently,
ab+bc+ca+2(a+b+c)+3≥abc+ab+bc+ca+a+b+c+1. ab + bc + ca + 2(a + b + c) + 3 \ge abc + ab + bc + ca + a + b + c + 1.
Applying the assumption a+b+c=6a+b+c=6 and collecting similar terms reduces this inequality to 8≥abc8 \ge abc. By AM-GM, 2=63=a+b+c3≥abc32 = \frac{6}{3} = \frac{a+b+c}{3} \ge \sqrt[3]{abc}. Hence 8=23≥abc8 = 2^3 \ge abc, completing the proof.

Solution 3

As in Solution 1, we reduce the problem to the inequality (12). The latter is equivalent to the inequality
1a+1+1b+1+1c+13≥13, \frac{\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}}{3} \ge \frac{1}{3},
By AM-HM,
31a+1+1b+1+1c+1≤(a+1)+(b+1)+(c+1)3=a+b+c+33. \frac{3}{\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}} \le \frac{(a+1) + (b+1) + (c+1)}{3} = \frac{a+b+c+3}{3}.
As a+b+c=6a+b+c=6, we obtain a+b+c+33=3\frac{a+b+c+3}{3} = 3, completing the proof.

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