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.

On Vacca's formula

One thought on “Huygen’s Inequality”

Two more inequalities:

1. For $\alpha \in \left(0, \frac{\pi}{2} \right)$ it holds that

$\alpha > \frac{3\sin \alpha}{2 +\cos \alpha}$

2. For $x \geq0$ it holds that

$2 \sinh x + \tanh x \geq 3x$