Let x be a positive acute angle. Prove that cos2(x)cot(x)+sin2(x)tan(x)≥1
Solution
The geometric-arithmetic inequality gives cosxsinx≤2cos2x+sin2x=21. It follows that 1=(cos2x+sin2x)2=cos4x+sin4x+2cos2xsin2x≤cos4x+sin4x+21 so cos4x+sin4x≥21≥cosxsinx. The required inequality follows.
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Source: MathNet,
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