Estonian Mathematical Olympiad

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

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Alternative solution.

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 $$\frac{ab+bc+ca+2(a+b+c)+3}{abc+ab+bc+ca+a+b+c+1}\ge 1,$$ or equivalently, $$ab+bc+ca+2(a+b+c)+3\ge abc+ab+bc+ca+a+b+c+1.$$ Applying the assumption $a+b+c=6$ and collecting similar terms reduces this inequality to $8\ge abc$. By AM-GM, $2=\frac{6}{3}=\frac{a+b+c}{3}\ge \sqrt[3]{abc}$. Hence $8=2^3\ge abc$, completing the proof.

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Alternative solution.

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