An Arctangent Mean Bound

★ 7/10trigonometry calculus Substitution Monotonicity inequality

Problem

For $0 < a < b$ , prove that

\frac{\arctan b - \arctan a}{b - a} > \frac{1}{\sqrt{1+a^2}\cdot\sqrt{1+b^2}}.

Answer

Solution

Difficulty	7/10
Topics	trigonometry, calculus, Substitution, Monotonicity, inequality

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