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Baltic Way 1995 · Problem 6

Algebra

Prove that for positive a,b,c,da, b, c, d

a+ca+b+b+db+c+c+ac+d+d+bd+a≥4.\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a} \geq 4 .
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Review

Topics

Equations and inequalities

Solutions

Solution

Solution: The inequality between the arithmetic and harmonic mean gives

a+ca+b+c+ac+d≥4a+ba+c+c+dc+a=4⋅a+ca+b+c+db+db+c+d+bd+a≥4b+cb+d+d+ad+b=4⋅b+da+b+c+d\begin{aligned} & \frac{a+c}{a+b}+\frac{c+a}{c+d} \geq \frac{4}{\frac{a+b}{a+c}+\frac{c+d}{c+a}} = 4 \cdot \frac{a+c}{a+b+c+d} \\ & \frac{b+d}{b+c}+\frac{d+b}{d+a} \geq \frac{4}{\frac{b+c}{b+d}+\frac{d+a}{d+b}} = 4 \cdot \frac{b+d}{a+b+c+d} \end{aligned}

and adding these inequalities yields the required inequality.

Contest context

Results from Baltic Way 1995

9 teams

Mean score
2.8 / 5
Scores of 4 or 5
4 / 9
Estonia
0 / 5

Score distribution

03
10
21
31
40
54
All team scores
TeamScore
Poland5 / 5
Latvia5 / 5
Sweden5 / 5
Lithuania5 / 5
Denmark0 / 5
Finland0 / 5
St. Petersburg3 / 5
Estonia0 / 5
Iceland2 / 5