MHB Is Sine of x Greater Than or Equal to 2x Over Pi for Angles Up to Pi/2?

[anemone]

Here is this week's POTW:

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Let $0\le x \le \dfrac{\pi}{2}$. Prove that $\sin x \ge \dfrac{2x}{\pi}$.

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[anemone]

Congratulations to lfdahl for his correct solution (Cool) , which you can find below:

Spoiler

Since $\sin’’(x) = -\sin x \leq 0$ for $x \in [0; \frac{\pi}{2}]$, the function $\sin x$ is concave (downward).

Per definition any point on any secant to the graph lies below the graph (and on the graph at the end points).

But $y = \frac{2}{\pi}x$ is a secant to $\sin x$ on the given interval. Therefore $\sin x \geq \frac{2x}{\pi}$ for $x \in [0; \frac{\pi}{2}]$.