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Prove that

$$1- \frac{x}{3} \le \frac{\sin x}x \le 1.1 - \frac{x}{4}, \quad \forall x\in(0,\pi].$$

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From the concavity of $f(x)=\cos x$ over $[0,\pi/2]$, we have:

$$ \forall x\in[0,\pi/2],\quad \cos x\geq 1-\frac{2}{\pi}x, $$

from which

$$ \forall x\in[0,\pi/2],\quad \sin x\geq \frac{1}{\pi}x(\pi-x) = x-\frac{1}{\pi}x^2$$

follows, by integration. Now, both the RHS and the LHS are symmetric wrt $x=\frac{\pi}{2}$, so we can extend the inequality over the whole $[0,\pi]$ interval:

$$ \forall x\in[0,\pi],\quad \frac{\sin x}{x}\geq 1-\frac{x}{\pi}\geq 1-\frac{x}{3}, $$

proving the "easier" inequality. Using an analogue tecnique, it's possible to establish something a little weaker than the other desired inequality. Since:

$$\operatorname{argmax}_{[0,\pi]}\left(\cos x+\frac{x}{2}\right)=\frac{\pi}{6},$$

we have:

$$ \forall x\in[0,\pi],\quad \cos x+\frac{1}{2}-\left(\frac{\pi}{12}+\cos\frac{\pi}{6}\right)\leq 0,$$

so, by integration,

$$\forall x\in[0,\pi],\quad \sin x +\frac{x^2}{4} - \left(\frac{\pi}{12}+\cos\frac{\pi}{6}\right)x \leq 0,$$

holds, that is:

$$\forall x\in[0,\pi],\quad \frac{\sin x}{x}\leq 1.1278247\ldots-\frac{x}{4}.$$

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