Proving $\sin 2x \geq x$, for $x \in \left[0, \frac{\pi}{4}\right]$

Question

I want to prove the following inequality:

$$\sin 2x \geq x \;\;\text{for}\;\; x \in \left[0, \frac{\pi}{4}\right]$$

I know that $\sin 2x = 2\sin x\cos x$, and I tried to use the Taylor series, but I could't prove the above inequality. Any ideas?

Answer 1

I presume you actually want $x\le\sin 2x$ for $0\le x\le \pi/4$.

If $$f(x)=\frac{\sin x}{x}$$ then $$f(x)=\int_0^1\cos xt\,dt.$$ On the interval $(0,\pi/2)$ therefore $f$ is decreasing. In particular on this interval $$f(x)\ge f(\pi/2)=\frac 2\pi>\frac12$$ so that $\sin x\ge x/2$ for $0\le x\le \pi/2$.

Answer 2

Let $f(x)=\sin2x-x$.

Thus, $$f''(x)=-4\sin2x\leq0,$$ which says that $f$ is a concave function.

Id est, $$\min\limits_{\left[0,\frac{\pi}{4}\right]}f=\min\left\{f(0),f\left(\frac{\pi}{4}\right)\right\}=f(0)=0.$$ Thus, indeed, $\sin2x\geq x.$