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Baltic Way 2010 · Problem 2

Algebra

Let xx be a real number such that 0<x<π20<x<\frac{\pi}{2}. Prove that

cos⁡2(x)cot⁡(x)+sin⁡2(x)tan⁡(x)≥1. \cos ^{2}(x) \cot (x)+\sin ^{2}(x) \tan (x) \geq 1 \text {. }
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Review

Topics

Equations and inequalities

Solutions

Solution

The geometric-arithmetic inequality gives

cos⁡xsin⁡x≤cos⁡2x+sin⁡2x2=12.\cos x \sin x \le \frac{\cos^2 x + \sin^2 x}{2} = \frac{1}{2}.

It follows that

1=(cos⁡2x+sin⁡2x)2=cos⁡4x+sin⁡4x+2cos⁡2xsin⁡2x≤cos⁡4x+sin⁡4x+121 = (\cos^2 x + \sin^2 x)^2 = \cos^4 x + \sin^4 x + 2 \cos^2 x \sin^2 x \le \cos^4 x + \sin^4 x + \frac{1}{2}

so

cos⁡4x+sin⁡4x≥12≥cos⁡xsin⁡x.\cos^4 x + \sin^4 x \ge \frac{1}{2} \ge \cos x \sin x.

The required inequality follows.

Contest context

Results from Baltic Way 2010

10 teams

Mean score
5.0 / 5
Scores of 4 or 5
10 / 10
Estonia
5 / 5

Score distribution

00
10
20
30
40
510
All team scores
TeamScore
Poland5 / 5
Lithuania5 / 5
Germany5 / 5
Latvia5 / 5
Denmark5 / 5
Sweden5 / 5
Estonia5 / 5
Norway5 / 5
Finland5 / 5
Iceland5 / 5