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Huygen’s Inequality

Let \alpha \in \left[0, \frac{\pi}{2} \right). Prove that

    \[2 \sin \alpha + \tan \alpha \geq 3\alpha\]

Solution

Let f(\alpha) = 2 \sin \alpha + \tan \alpha. f is twice differentiable with

    \[f''(\alpha) = -2 \sin \alpha +2 \tan \alpha + 2 \tan^3 \alpha \geq 0\]

Hence f is convex. The tangent at (0, 0) has equation y=3x. The result follows.

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